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Work
Work
Work is the product of the magnitude of the __________________ moved times the component of a ________________ in the direction of the ________________.
Work
Work is the product of the magnitude of the displacement moved times the component of a force in the direction of the displacement.
Work
Work is the product of the magnitude of the displacement moved times the component of a force in the direction of the displacement.
F
Δd
| F | cosθ
θ
Work
Work is the product of the magnitude of the displacement moved times the component of a force in the direction of the displacement.
Defining equation:
?
F
Δd
| F | cosθ
θ
Work
Work is the product of the magnitude of the displacement moved times the component of a force in the direction of the displacement.
Defining equation:
W = | Δd | X | F | cosθ
where θ is the smallest angle
between Δd and F
in a “ tail-to-tail “ diagram
F
Δd
| F | cosθ
θ
Work
W = | Δd | X | F | cosθ
W can be written using a special mathematical “vector operator” How?
Work
W = | Δd | X | F | cosθ
W can be written using a special mathematical “vector operator”
W = Δd ● F
Work
W = | Δd | X | F | cosθ
W can be written using a special mathematical “vector operator”
W = Δd ● F note that ● stands for “dot” or “scalar”
product
Work
W = | Δd | X | F | cosθ
W can be written using a special mathematical “vector operator”
W = Δd ● F note that ● stands for “dot” or “scalar”
product
So Δd ● F = | Δd | X | F | cosθ
Work
W = | Δd | X | F | cosθ
W can be written using a special mathematical “vector operator”
W = Δd ● F note that ● stands for “dot” or “scalar”
product
So Δd ● F = | Δd | X | F | cosθ
Since Work is a product of magnitudes only, it is a _____ quantity.
Work
W = | Δd | X | F | cosθ
W can be written using a special mathematical “vector operator”
W = Δd ● F note that ● stands for “dot” or “scalar”
product
So Δd ● F = | Δd | X | F | cosθ
Since Work is a product of magnitudes only, it is a scalar quantity.
Work
So Work can be defined mathematically as...
W = Δd ● F or | Δd | X | F | cosθ
Work
So Work can be defined mathematically as...
W = Δd ● F or | Δd | X | F | cosθ What is the SI unit of work?
Work
So Work can be defined mathematically as...
W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = ?
Work
So Work can be defined mathematically as...
W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = N m
= ?
Work
So Work can be defined mathematically as...
W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = N m
= ( kg m /s2 ) m
= ?
Work
So Work can be defined mathematically as...
W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = N m
= ( kg m /s2 ) m
= kg m2 / s2
Work
So Work can be defined mathematically as...
W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = N m
= ( kg m /s2 ) m
= kg m2 / s2
Note that “F” in the formula can be any kind of force
Work
So Work can be defined mathematically as...
W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = N m
= ( kg m /s2 ) m
= kg m2 / s2
Note that “F” in the formula can be any kind of force Both the force doing the work and the object that the
work is being done on must be specified or understood.
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 N
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Na) What is the
work done on the box by the applied force?
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Na) What is the
work done on the box by the applied force?
Draw a tail-to-tail diagram
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Na) What is the
work done on the box by the applied force?
W = | Δd | X | F | cosθ
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Na) What is the
work done on the box by the applied force?
W = | Δd | X | F | cosθ = 5.00 m X 100.0N cos36.9°
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Na) What is the
work done on the box by the applied force?
W = | Δd | X | F | cosθ = 5.00 m X 100.0N cos36.9° = 400 J
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Fa = 100 N
θ = 36.9°
Δd = 5.00 m
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nb) What is the
work done on the box by the net force?
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nb) What is the
work done on the box by the net force?
Find the net force. It acts along the horizontal-axis.
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nb) What is the
work done on the box by the net force?
Fnet
= 100 cos 36.9° - 22 = ?
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nb) What is the
work done on the box by the net force?
Fnet
= 100 cos 36.9° - 22 = 58.0 N [E]
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nb) What is the
work done on the box by the net force?
Draw a tail-to-tail diagram
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nb) What is the
work done on the box by the net force?
Draw a tail-to-tail diagram Fnet
= 58.0 N
Δd = 5.00 m
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nb) What is the
work done on the box by the net force?
Fnet
= 58.0 N
Δd = 5.00 m
W = | Δd | X | F | cosθ
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nb) What is the
work done on the box by the net force?
Fnet
= 58.0 N
Δd = 5.00 m
W = | Δd | X | F | cosθ = 5.00 m X 58.0 N cos 0°
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nb) What is the
work done on the box by the net force?
Fnet
= 58.0 N
Δd = 5.00 m
W = | Δd | X | F | cosθ = 5.00 m X 58.0 N cos 0° = 290 J
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nb) What is the
work done on the box by the net force?
Fnet
= 58.0 N
Δd = 5.00 m
W = | Δd | X | F | cosθ = 5.00 m X 58.0 N cos 0° = 290 J
Note that if θ = 0, then a short-cut formula can be used W = F d
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nb) What is the
work done on the box by the net force?
Fnet
= 58.0 N
Δd = 5.00 m
W = | Δd | X | F | cosθ = 5.00 m X 58.0 N cos 0° = 290 J
Note that if θ = 0, then a short-cut formula can be used W = F d
The work done by the net force is sometimes called the “net work done” or W
net
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nc) What is the
work done on the box by the kinetic friction force?
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nc) What is the
work done on the box by the kinetic friction force?
Draw a tail-to-tail diagram
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nc) What is the
work done on the box by the kinetic friction force?
Draw a tail-to-tail diagram
fK = 22.0 N Δd = 5.00 m
θ = ?
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nc) What is the
work done on the box by the kinetic friction force?
Draw a tail-to-tail diagram
fK = 22.0 N Δd = 5.00 m
θ = 180°
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nc) What is the
work done on the box by the kinetic friction force?
fK = 22.0 N Δd = 5.00 m
θ = 180°W = | Δd | X | F | cosθ = ?
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nc) What is the
work done on the box by the kinetic friction force?
fK = 22.0 N Δd = 5.00 m
θ = 180°W = | Δd | X | F | cosθ =5.00 m X 22 N cos 180°
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nc) What is the
work done on the box by the kinetic friction force?
fK = 22.0 N Δd = 5.00 m
θ = 180°W = | Δd | X | F | cosθ =5.00 m X 22 N cos 180° = - 110 J
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nc) What is the
work done on the box by the kinetic friction force?
fK = 22.0 N Δd = 5.00 m
θ = 180°W = | Δd | X | F | cosθ =5.00 m X 22 N cos 180° = - 110 J
The negative sign means energy is lost to the surroundings as heat.
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nc) What is the
work done on the box by the kinetic friction force?
fK = 22.0 N Δd = 5.00 m
θ = 180°W = | Δd | X | F | cosθ =5.00 m X 22 N cos 180° = - 110 J
The negative sign means energy is lost to the surroundings as heat.
Note that fk has θ = 180°,
so a short-cut formula can be used W = - f
k d
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nd) What is the
work done on the box by the weight?
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nd) What is the
work done on the box by the weight?
Draw a tail-to-tail diagram
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nd) What is the
work done on the box by the weight?
Draw a tail-to-tail diagram F
g = 480 N
Δd = 5.00 m
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nd) What is the
work done on the box by the weight?
Draw a tail-to-tail diagram F
g = 480 N
Δd = 5.00 m
θ = ?
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nd) What is the
work done on the box by the weight?
Draw a tail-to-tail diagram F
g = 480 N
Δd = 5.00 m
θ = 90°
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nd) What is the
work done on the box by the weight?
Fg = 480 N
Δd = 5.00 m
θ = 90°W = | Δd | X | F | cosθ = ?
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nd) What is the
work done on the box by the weight?
Fg = 480 N
Δd = 5.00 m
θ = 90°W = | Δd | X | F | cosθ = 5.00 m X 480N cos 90° = ?
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nd) What is the
work done on the box by the weight?
Fg = 480 N
Δd = 5.00 m
θ = 90°W = | Δd | X | F | cosθ = 5.00 m X 480N cos 90° = 0 J
Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of 100.0 N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N.
Rough floor
m = 48.0 kgF
a = 100 N
θ = 36.9°
Δd = 5.00 m
fK = 22.0 Nd) What is the
work done on the box by the weight?
Fg = 480 N
Δd = 5.00 m
θ = 90°W = | Δd | X | F | cosθ = 5.00 m X 480N cos 90° = 0 J
Note that if θ = 90°, no work is done
Conditions where no work is done
Conditions where no work is done
1. If θ = ?
Conditions where no work is done
1. If θ = 90°
Conditions where no work is done
1. If θ = 90°
2. If | F | = ?
Conditions where no work is done
1. If θ = 90°
2. If | F | = 0 N
Conditions where no work is done
1. If θ = 90°
2. If | F | = 0 N
3. If | Δd | = ?
Conditions where no work is done
1. If θ = 90°
2. If | F | = 0 N
3. If | Δd | = 0 m
Energy
Energy
Definition in words: ?
Energy
Energy is the ability to do work
Energy
Energy is the ability to do work Since energy is defined in terms of “work”, it is a
__________ quantity.
Energy
Energy is the ability to do work Since energy is defined in terms of “work”, it is a
scalar quantity.
Energy
Energy is the ability to do work Since energy is defined in terms of “work”, it is a
scalar quantity. Like work, the SI unit for any type of energy is the
__________.
Energy
Energy is the ability to do work Since energy is defined in terms of “work”, it is a
scalar quantity. Like work, the SI unit for any type of energy is the
Joule.
Energy of Motion
Energy of Motion
Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies.
Energy of Motion
Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies.
Energy that an object possesses by virtue of its motion is called ____________ energy.
Energy of Motion
Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies.
Energy that an object possesses by virtue of its motion is called kinetic energy.
Energy of Motion
Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies.
Energy that an object possesses by virtue of its motion is called kinetic energy.
The symbol for kinetic energy is ______.
Energy of Motion
Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies.
Energy that an object possesses by virtue of its motion is called kinetic energy.
The symbol for kinetic energy is Ek.
Energy of Motion
Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies.
Energy that an object possesses by virtue of its motion is called kinetic energy.
The symbol for kinetic energy is Ek.
What is the formula for kinetic energy?
Energy of Motion
Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies.
Energy that an object possesses by virtue of its motion is called kinetic energy.
The symbol for kinetic energy is Ek.
Ek = mv2/2
Try this example: Compare the kinetic energy of a 5055 kg truck and a 20 g bee moving at 36.0 km/h. E
k = mv2/2
Try this example: Compare the kinetic energy of a 5055 kg truck and a 20 g bee moving at 36.0 km/h. E
k = mv2/2
Truck
m = 5055 kg
v = 36.0 km/h = 10.0 m/s
Ek = ?
Ek = mv2/2 = 5055(10.0)2/2
= 2.5 X 105 J
Bee
m = 0.020 kg
v = 36.0 km/h = 10.0 m/s
Ek = ?
Ek = mv2/2 = 0.020(10.0)2/2
= 1.0 J
Try this example #2 : An 8.00 kg bird has a kinetic energy of 576 J. How fast is it moving?
Try this example #2 : An 8.00 kg bird has a kinetic energy of 576 J. How fast is it moving?
Given: m = 8.00 kg Ek = 576 J
Unknown: v = ?
Formula: Ek = mv2/2 so v = (2 E
k / m )1/2
Sub: v = (2 X 576 / 8.00 )1/2
= 12 m/s
Energy of raised position
Energy of raised position
An object that is raised above a certain reference level “is able” or has “potential” to do work.
Energy of raised position
An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it.
Energy of raised position
An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it.
The energy of raised position is called _________ __________ energy.
Energy of raised position
An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it.
The energy of raised position is called gravitational potential energy. Its symbol is ____.
Energy of raised position
An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it.
The energy of raised position is called gravitational potential energy. Its symbol is E
g.
Energy of raised position
An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it.
The energy of raised position is called gravitational potential energy. Its symbol is E
g.
The formula for gravitational potential energy is
___________
Energy of raised position
An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it.
The energy of raised position is called gravitational potential energy. Its symbol is E
g.
The formula for gravitational potential energy is
Eg = mgh
More About the formula for gravitational potential energy
More About the formula for gravitational potential energy
Eg = mgh
More About the formula for gravitational potential energy
Eg = mgh
Note that the height or “h” and therefore Eg depends
on the reference level used to specify Eg = 0 or the
zero level of gravitational potential energy.
More About the formula for gravitational potential energy
Eg = mgh
Note that the height or “h” and therefore Eg depends
on the reference level used to specify Eg = 0 or the
zero level of gravitational potential energy. Usually “ground level” is the specified zero reference
level but not always. The zero reference level could be the top of a table or the bottom of the swing of a pendulum.
More About the formula for gravitational potential energy
Eg = mgh
Note that the height or “h” and therefore Eg depends
on the reference level used to specify Eg = 0 or the
zero level of gravitational potential energy. Usually “ground level” is the specified zero reference
level but not always. The zero reference level could be the top of a table or the bottom of the swing of a pendulum.
Note “h” can be + or – depending on whether it is “above' or “below” the zero reference level.
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= ?
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0 = ?
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0 = 840 J
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0 = 840 J
b) Eg = mgh
= ?
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0 = 840 J
b) Eg = mgh
= 14 X 10.0 ( -4 )
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0 = 840 J
b) Eg = mgh
= 14 X 10.0 ( -4 ) = ?
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0 = 840 J
b) Eg = mgh
= 14 X 10.0 ( -4 ) = - 560 J
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0 = 840 J
b) Eg = mgh
= 14 X 10.0 ( -4 ) = - 560 J
c) Eg = mgh
= ?
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0 = 840 J
b) Eg = mgh
= 14 X 10.0 ( -4 ) = - 560 J
c) Eg = mgh
= 14 X 10.0 (0)
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0 = 840 J
b) Eg = mgh
= 14 X 10.0 ( -4 ) = - 560 J
c) Eg = mgh
= 14 X 10.0 (0) = ?
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0 = 840 J
b) Eg = mgh
= 14 X 10.0 ( -4 ) = - 560 J
c) Eg = mgh
= 14 X 10.0 (0) = 0 J
Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge
m = 14 kg
ground
ledge
Cliff top
4.0 m
6.0 m
a) Eg = mgh
= 14 X 10.0 X 6.0 = 840 J
b) Eg = mgh
= 14 X 10.0 ( -4 ) = - 560 J
c) Eg = mgh
= 14 X 10.0 (0) = 0 J
Temperature is another scalar quantity like E
g and it can have negative values
too. It just depends on the zero reference level as well. The freezing point of water is 0° C in the Celsius scale but 273° K in the Kelvin scale.
Energy stored in a stretched or compressed spring
Energy stored in a stretched or compressed spring
Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy.
Energy stored in a stretched or compressed spring
Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy.
The energy of a deformed spring is called ___________ ___________ energy.
Energy stored in a stretched or compressed spring
Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy.
The energy of a deformed spring is called elastic potential energy.
Energy stored in a stretched or compressed spring
Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy.
The energy of a deformed spring is called elastic potential energy.
The symbol for elastic potential energy is _____.
Energy stored in a stretched or compressed spring
Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy.
The energy of a deformed spring is called elastic potential energy.
The symbol for elastic potential energy is Ee.
Energy stored in a stretched or compressed spring
Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy.
The energy of a deformed spring is called elastic potential energy.
The symbol for elastic potential energy is Ee.
To understand further, we need to know about forces on deformed springs.
Forces needed to “deform” a spring
Forces needed to “deform” a springEquilibrium or rest position
stretched
X
X
compression
Fs
Fs
Forces needed to “deform” a springEquilibrium or rest position
stretched
X
X
compression
Fs
Fs
X = amount of compression or stretch of a spring from its rest position (deformation)
Forces needed to “deform” a springEquilibrium or rest position
stretched
X
X
compression
Fs
Fs
X = amount of compression or stretch of a spring from its rest position (deformation)
Fs = magnitude of force
needed to deform a spring
Forces needed to “deform” a springEquilibrium or rest position
stretched
X
X
compression
Fs
Fs
X = amount of compression or stretch of a spring from its rest position (deformation)
Fs = magnitude of force
needed to deform a spring
Robert Hooke discovered that F
s α X
Hooke's law
More on Hooke's Law
More on Hooke's Law
Fs α X
More on Hooke's Law
Fs α X How can we change this to an equation?
More on Hooke's Law
Fs α X
Fs = KX
More on Hooke's Law
Fs α X
Fs = KX
K , the constant of proportionality, is called the __________ constant of the spring or
__________ constant of the spring or just
the _________ constant.
More on Hooke's Law
Fs α X
Fs = KX
K , the constant of proportionality, is called the force constant of the spring or
__________ constant of the spring or just
the _________ constant.
More on Hooke's Law
Fs α X
Fs = KX
K , the constant of proportionality, is called the force constant of the spring or
Hooke's constant of the spring or just
the _________ constant.
More on Hooke's Law
Fs α X
Fs = KX
K , the constant of proportionality, is called the force constant of the spring or
Hooke's constant of the spring or just
the spring constant.
More on Hooke's Law
Fs α X
Fs = KX
K , the constant of proportionality, is called the force constant of the spring or
Hooke's constant of the spring or just
the spring constant.
What does K depend on?
More on Hooke's Law
Fs α X
Fs = KX
K , the constant of proportionality, is called the force constant of the spring or
Hooke's constant of the spring or just
the spring constant.
K depends on the “stiffness” of the spring involved
More on Hooke's Law
Fs α X
Fs = KX
K , the constant of proportionality, is called the force constant of the spring or
Hooke's constant of the spring or just
the spring constant.
K depends on the “stiffness” of the spring involved
If Fs = KX , then K = ?
More on Hooke's Law
Fs α X
Fs = KX
K , the constant of proportionality, is called the force constant of the spring or
Hooke's constant of the spring or just
the spring constant.
K depends on the “stiffness” of the spring involved
If Fs = KX , then K = Fs / X
More on Hooke's Law
Fs α X
Fs = KX
K , the constant of proportionality, is called the force constant of the spring or
Hooke's constant of the spring or just
the spring constant.
K depends on the “stiffness” of the spring involved
If Fs = KX , then K = Fs / X and SI unit for K is ______.
More on Hooke's Law
Fs α X
Fs = KX
K , the constant of proportionality, is called the force constant of the spring or
Hooke's constant of the spring or just
the spring constant.
K depends on the “stiffness” of the spring involved
If Fs = KX , then K = Fs / X and SI unit for K is N/m.
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Draw an FBD of the mass
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Fs
Fg
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Fnety
= ?
Fs
Fg
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Fnety
= 0
Fs
Fg
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Fnety
= 0 F
s + F
g = 0
Fs
Fg
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Fnety
= 0 F
s + F
g = 0
+KX – 5.00 = 0
Fs
Fg
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Fnety
= 0 F
s + F
g = 0
+KX – 5.00 = 0 +K(.0800) – 5.00 = 0
Fs
Fg
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Fnety
= 0 F
s + F
g = 0
+KX – 5.00 = 0 +K(.0800) – 5.00 = 0 K(.0800) = 5.00
Fs
Fg
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Fnety
= 0 F
s + F
g = 0
+KX – 5.00 = 0 +K(.0800) – 5.00 = 0 K(.0800) = 5.00 K= 5.00 / .0800
Fs
Fg
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Fnety
= 0 F
s + F
g = 0
+KX – 5.00 = 0 +K(.0800) – 5.00 = 0 K(.0800) = 5.00 K= 5.00 / .0800 K= 62.5 N/m
Fs
Fg
Hooke's Law Example: A 0.500 kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring.
X = 8.00 cmm =0.500 kg
rest
Fnety
= 0 F
s + F
g = 0
+KX – 5.00 = 0 +K(.0800) – 5.00 = 0 K(.0800) = 5.00 K= 5.00 / .0800 K= 62.5 N/m
Fs
Fg
This k value tells us that this spring requires 62.5 N of force to stretch or compress it by 1 meter. However, most strings have a threshold limit of deformation, beyond which Hooke's law does not hold.
Elastic potential energy
Elastic potential energy
Let's plot Fs vs X
Fs
X
Elastic potential energy
Let's plot Fs vs X
Fs
X
Elastic potential energy
Let's plot Fs vs X
Fs
X
The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve.
Elastic potential energy
Let's plot Fs vs X
Fs
X
The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the
spring is stretched X meters as shown
Elastic potential energy
Let's plot Fs vs X
Fs
X
The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the
spring is stretched X meters as shown
In terms of X, what isF
s = ?
Elastic potential energy
Let's plot Fs vs X
Fs
X
The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the
spring is stretched X meters as shown
In terms of X,F
s = KX
Elastic potential energy
Let's plot Fs vs X
Fs
X
The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the
spring is stretched X meters as shown
In terms of X,F
s = KX
Elastic potential energy, Ee= the
area under the curve. The area is a triangle. What is the formula for the area of a triangle?
Elastic potential energy
Let's plot Fs vs X
Fs
X
The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the
spring is stretched X meters as shown
In terms of X,F
s = KX
Elastic potential energy, Ee= the
area under the curve. The area is a triangle. A = b h / 2
Elastic potential energy
Let's plot Fs vs X
Fs
X
The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the
spring is stretched X meters as shown
In terms of X,F
s = KX
Elastic potential energy, Ee= the
area under the curve. The area is a triangle. A = b h / 2Therefore E
e = ?
Elastic potential energy
Let's plot Fs vs X
Fs
X
The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the
spring is stretched X meters as shown
In terms of X,F
s = KX
Elastic potential energy, Ee= the
area under the curve. The area is a triangle. A = b h / 2Therefore E
e = (X ) ( KX ) / 2
Elastic potential energy
Let's plot Fs vs X
Fs
X
The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the
spring is stretched X meters as shown
In terms of X,F
s = KX
Elastic potential energy, Ee= the
area under the curve. The area is a triangle. A = b h / 2Therefore E
e = (X ) ( KX ) / 2
Ee
= KX2 / 2
Elastic potential energy
Let's plot Fs vs X
Fs
X
The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the
spring is stretched X meters as shown
In terms of X,F
s = KX
Elastic potential energy, Ee= the
area under the curve. The area is a triangle. A = b h / 2Therefore E
e = (X ) ( KX ) / 2
Ee
= KX2 / 2
The elastic potential energy formula is...
Ee
= KX2 / 2
Try this! The spring in the previous example has a force constant of 62.5 N/m and is stretched 8.00 cm. How much elastic potential energy does this stretched spring have?
Try this! The spring in the previous example has a force constant of 62.5 N/m and is stretched 8.00 cm. How much elastic potential energy does this stretched spring have?
Given: K = 62.5 N/m X = 8.00 cm or 0.0800 m
Unknown: Ee = ?
Formula: Ee = KX2 / 2
Sub: Ee = (62.5)(0.0800)2 / 2
= 0.200 J Answer: The spring has 0.200 J of elastic potential
energy. It is able to do 0.200 J of work.
