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It’s “Moving Day”…. Dave tries to push a refrigerator over thick carpet . He pushes hard, but it doesn’t budge. Steve pushes an identical refrigerator across a wooden floor . He also pushes hard, and it moves easily. Who does more “work”? Stev e

It’s “Moving Day”…

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It’s “Moving Day”…. Dave tries to push a refrigerator over thick carpet . He pushes hard, but it doesn’t budge. Steve pushes an identical refrigerator across a wooden floor . He also pushes hard, and it moves easily. Who does more “work”?. Steve. - PowerPoint PPT Presentation

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Page 1: It’s “Moving Day”…

It’s “Moving Day”….

Dave tries to push a refrigerator over thick carpet. He pushes hard, but it doesn’t budge.

Steve pushes an identical refrigerator across a wooden floor. He also pushes hard, and it moves easily.

Who does more “work”? Steve

Page 2: It’s “Moving Day”…

Steve moved the refrigerator….Dave didn’t.

Work – what is accomplished when a force acts on an object and that force causes the object to move over a certain distance

W = FdcosΘ

W = Work (Joule)F = force (N)

d = distance (m)Θ= angle between

the force and distance

Page 3: It’s “Moving Day”…

Is work being done by the man on the groceries by holding it in place?

Is work being done by the Earth on the groceries?

The net work in the vertical direction on the groceries is zero (net force = 0)

Is work being done by the man on the groceries if he walks forward?

No…his force on the bag is perpendicular to motion

Page 4: It’s “Moving Day”…

Work is a scalar quantity -- it only has magnitude. However, there is a difference between negative work and positive work.

The hammer does positive work on the nail (same direction as displacement)

The nail does negative work on the hammer (opposite direction of displacement)

Page 5: It’s “Moving Day”…

A person pulls a 50-kg crate 40 m along a horizontal floor with a constant force of 100 N at an angle of 37 . The floor is rough and exerts a friction force of 50 N. Determine the work done by each force acting on the crate and the net work being done on the crate.

Page 6: It’s “Moving Day”…

If a net work is done on the bus above, what do we know about the bus

•It moves

•A force is applied

•It accelerates

If an object’s speed changes, its kinetic energy changes

Kinetic energy – energy of motion (measured in joules)

Page 7: It’s “Moving Day”…

KE = 1/2 mv2

KE = kinetic energy (J)

m = mass (kg)

v = speed (m/s)

Since work causes an object’s speed to change, it causes its kinetic energy to change

(Work-Energy Theorem)

ΣW = ∆KE

Page 8: It’s “Moving Day”…

ΣW = 1/2 mvf2 -1/2 mvi

2

If ΣW (Wnet) > 0, KE increases (object speeds up)

If ΣW (Wnet) < 0, KE decreases (object slows down)

Page 9: It’s “Moving Day”…

A baseball (m = 140 g) traveling 32 m/s moves a fielder’s glove backward 25 cm when the ball is caught. What was the average force exerted by the ball on the glove?

Page 10: It’s “Moving Day”…

Power

If you press a barbell over your head with a constant speed….

Are you doing work on the barbell?

YesIs gravity doing work on the barbell?

Yes (negative)What is the net work on the barbell?

Zero (no a, no ΣF)

If you faster reps, you feel more fatigued. Why?

Your rate of doing work is increasing.

Page 11: It’s “Moving Day”…

Power = work / time

P = W / tP = power (Watt)

W = work (J)

T = time (s)

1 Horsepower = 550 ft lb/s = 746 W

Page 12: It’s “Moving Day”…

P = F d / t

P = F vP = power (W)

F = force (N)

v = average speed (m/s)

Since we know W = Fd,

Page 13: It’s “Moving Day”…

A pump is to lift 18.0 kg of water per minute to a height of 3.60 m. Assuming the pump moves the water at a constant speed, what power rating (watts) should the pump motor have?

Page 14: It’s “Moving Day”…

A car is traveling up a 10° incline at a constant speed of 80 km/h. The car has a mass of 1400 kg and the frictional force of the road on the car is 700 N. How much horesepower does the motor need in order to move at that speed?

Page 15: It’s “Moving Day”…

Forms of Energy

Energy

Mechanical Nonmechanical

Kinetic Potential

Gravitational Elastic

Potential energy – energy associated with forces that depend on the position of an object relative to its surroundings

Page 16: It’s “Moving Day”…

Gravitational Potential EnergyAn object gains energy when work is done to it to lift it off the ground to a position above the ground.

W = Fadcosθ

If object is raised at a constant speed, Fa = Fg. Therefore,

Page 17: It’s “Moving Day”…

W = Fd = PEG= mgh

PEg = gravitational potential energy (J)

m = mass (kg)

g = gravitational acceleration (m/s2)

h = height (m)

Page 18: It’s “Moving Day”…

A roller coaster with a mass of 1000 kg travels up a 25 m hill and then back down 10 m. What is the potential energy at the top of the hill and at the ending position on the other side?

Page 19: It’s “Moving Day”…

Elastic Potential Energy

A spring gains energy when work is done to it to compress or stretch it from its natural position

Hooke’s Law

Fp = kx

Fs = -kx

k = spring constant (N/m)x = amount of

compression/tension (m)

Fp increases as a spring is compressed, so average force is F = ½(0 + kx) = ½ kx. Therefore,

Page 20: It’s “Moving Day”…

W = Fd = PES = ½kx2

PEs = elastic potential energy (J)

k = spring constant (N/m)

x = amount of compression/tension (m)

Page 21: It’s “Moving Day”…

A spring with a force constant of 5.2 N/m has a relaxed length of 2.45 m. When a mass is attached to the end of the spring and allowed to come to rest, the vertical length of the spring is 3.57 m. Calculate the elastic potential energy stored in the spring.

Page 22: It’s “Moving Day”…

Principle of Conservation of Mechanical Energy

If only conservative forces are acting , the total mechanical energy of a system neither increases nor decreases in any process. It stays constant – it is conserved.

Conservative forces – work done does not depend on path; just the initial and final positions (gravitational force, elastic force)

Nonconservative forces – work done depends on path taken (friction force, air resistance, applied force, tension force)

Page 23: It’s “Moving Day”…

E = PE + KE

An object’s initial E at time 1 is equal to an objects final E at time 2 assuming no nonconservative forces acting.

E1 = E2

PE1 + KE1 = PE2 + KE2

Page 24: It’s “Moving Day”…

If the original height of a rock is 3.0 m, calculate the rock’s speed when it has fallen 2.0 m and the speed right before it hits the ground.

Page 25: It’s “Moving Day”…

A dart of mass 0.100 kg is pressed against the spring of a toy dart gun. The spring (k = 250 N/m) is compressed 6.0 cm and released. If the dart detaches from the spring when the spring reaches its natural length, what speed does the dart acquire?

Page 26: It’s “Moving Day”…

Law of Conservation of Energy

The total energy is neither increased nor decreased in any process. Energy can be transformed from one for to another, and transferred from one object to another, but the total amount remains constant.

In a process, mechanical energy may be “lost” or “gained” due to the work that is done by nonconservative forces, but it simply is converted into or converted from another type of energy (heat, light, electric energy, etc.)

Page 27: It’s “Moving Day”…

WNC = ∆KE + ∆PE

The work done by nonconservative forces (to produce heat, light, etc.) is equal to the sum of the change in kinetic energy and the change in potential energy.

Page 28: It’s “Moving Day”…

You drop a 2.0-kg ball from a height of 2.0 m and it bounces back to a height of 1.5 m. How much energy is “lost” in the bounce? What is the ball’s speed at the instant it leaves the ground after the bounce?

Page 29: It’s “Moving Day”…

An SUV and a sports car are both moving 60 mph down the highway.

Which one has more kinetic energy?

Which one has more momentum?

Our intuition tells us that the SUV has more momentum because it is larger.

Page 30: It’s “Moving Day”…

Is there any way to give the sports car more momentum so that it eventually is the same as the SUV?

How would we make their kinetic energies equal?

Increase the speed of the sports car.

The same is true to make their momentums the same – increase the velocity of the sports car.

Momentum - the product of an object’s mass and velocity

Page 31: It’s “Moving Day”…

p = m x v

p = momentum (kg x m/s)m = mass (kg)

v = velocity (m/s)

Momentum is different from energy in that it is a vector quantity. Direction matters!

How would we change the car’s momentum?

Accelerate the car by applying more force.

Page 32: It’s “Moving Day”…

Σ F = ma

Σ F = m(∆v/∆t)

Σ F = (m∆v)/∆t

Σ F = ∆p/∆t

Newton’s Second Law – the net force applied to an object is equal to the rate of the change of momentum of an object

Page 33: It’s “Moving Day”…

A tennis ball with a mass of 0.060 kg leaves a tennis racket on the serve with a velocity of 55 m/s. If the racket is in contact with the ball for 0.004 seconds, what is the force that is exerted on the ball?

Page 34: It’s “Moving Day”…

Water leaves a hose at a rate of 1.5 kg/s with a velocity of 20 m/s. If the water is aimed at a car window which stops it, what is the force exerted by the car on the water?

Page 35: It’s “Moving Day”…

Impulse

When you participate in a water balloon toss, how do you catch a water balloon?

“Soft Hands”

Your catch is spread over over a larger amount of time.

If you were to abruptly catch the balloon, it would break.

Page 36: It’s “Moving Day”…

F = ∆p/∆t

F∆t = ∆p

To change an object’s momentum, if the time decreases, the force has to increase (or vice versa)

Impulse – the product of a force exerted on an object and the time over which the force acts

Page 37: It’s “Moving Day”…

A 12-kg hammer strikes a nail at a velocity of 8.5 m/s and comes to rest in a time interval of 8.0 ms. What is the impulse given to the nail? What is the average force acting on the nail?

Page 38: It’s “Moving Day”…

Law of Conservation of Momentum

The total momentum of an isolated system of objects remains constant.

In a system with numerous objects, the individual momentums of the objects may change as the objects interact with each other, but the sum of their momenta remains constant.

mAvA0 + mBvB0 = mAvA +mBvB

Page 39: It’s “Moving Day”…

Explosions

Initially, a system has no momentum. Then, some event happens causing objects to move in various directions. Regardless of the direction or motion, all of the resultant momentums have to cancel out to equal zero (to match the initial conditions).

Initial Momentum = Final Momentum

mv = mAvA + mBvB

Page 40: It’s “Moving Day”…

What is the recoil velocity of a 5.0-kg rifle that shoots a 0.020-kg bullet at a speed of 620 m/s?

Page 41: It’s “Moving Day”…

Elastic Collisons

Initially, at least one object of a system has momentum. It then hits another object causing each object to change its velocity. All of the momentums of the objects after the collision will equal the momentums of the objects before the collision.

Initial Momentum = Final Momentum

mAvA0 + mBvB0 = mAvA +mBvB

Page 42: It’s “Moving Day”…

Because the collision is elastic, not only is momentum conserved, but all mechanical energy is conserved.

½mAvA02 + ½mBvB0

2 = ½mAvA2 + ½mBvB

2

mAvA02 + mBvB0

2 = mAvA2 + mBvB

2

mAvA02 - mAvA

2 = mBvB2 - mBvB0

2

mA(vA02 - vA

2) = mB(vB2 - vB0

2)

mA(vA0 - vA)(vA0 + vA) = mB(vB

- vB0)(vB + vB0)

Page 43: It’s “Moving Day”…

mA(vA0 - vA)(vA0 + vA) = mB(vB

- vB0)(vB + vB0)

mAvA0 + mBvB0 = mAvA +mBvB

mAvA0 - mAvA = mBvB

- mBvB0

mA(vA0 - vA) = mB(vB

- vB0)

vA0 + vA = vB + vB0

vA0 - vB0 = vB – vA (for elastic collisions)

Back to momentum…

Combining both equations…

Page 44: It’s “Moving Day”…

A ball of mass 0.440 kg moving east with a speed of 2.20 m/s collides head-on with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the velocity of each ball after the collision?

Page 45: It’s “Moving Day”…

Inelastic Collisons

Inelastic collisions are ones where two separate objects collide and stick together moving together as one unit afterward. Momentum is conserved in these types of collisions.

Initial Momentum = Final Momentum

mAvA + mBvB = (mA + mB)v

Mechanical energy is not conserved in these types of collisions (deformation, heat, etc.)

Page 46: It’s “Moving Day”…

A bullet is fired vertically into a 1.40-kg block of wood at rest directly above it. If the bullet has a mass of 29.0 g and a speed of 510 m/s, how high will the block rise after the bullet becomes embedded in it?