Chapter 8Work & Kinetic Energy

DEHS 2011-12Physics 1

Energy

• Energy can be thought of as a quantity that reflects the state of an object

• There are many different forms of energy, we will study mechanical energy:– Kinetic Energy energy due to motion– Potential Energy energy due to position in a

“force field”

Work

• Work is the amount of energy that is given to or taken from an object/system– It is a scalar quantity has no direction

• If energy is added to work is positive• If energy is taken from work is negative

• It is measured in Joules (1 J = 1 kg.m2/s2)

Work done by a constant force

• Only valid when the force is constant, so you can’t use it when the force depends on the position (e.g. when attached to a spring)

• ϕ represents the smaller of the two angles between the Force and displacement vectors– It has a range: 0° ≤ ϕ ≤ 180°

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W = Fdcosφ

Work done by a constant force , special cases

• When the force is parallel to the displacement (ϕ=0°):

• When the force is antiparallel to the displacement (ϕ=180°):

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W=Fdcosφ€

W = Fdcos0°

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* W = Fd

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W = Fdcosφ

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W = Fdcos180°

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* W = −Fd

Work done by a constant force , special cases

• When the force is perpendicular to the displacement (ϕ=90°):

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W=Fdcosφ

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W = Fdcos90°

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* W = 0

F

d

No Work is done on this box!

Example 8-1An intern pushes a 72-kg patient on a 15-kg gurney, producing an

acceleration of 0.60 m/s2. How much work does the intern do by pushing the patient and gurney through a distance of 2.5 m?

Example 8-2In a gravity escape system (GES), an enclosed lifeboat on a large ship is

deployed by letting it slide down a ramp and then continuing in free fall to the water below. Suppose a 4970-kg lifeboat slides a distance of 5.00 m on a ramp, dropping through a vertical height of 2.50 m. How much work does gravity do on the boat?

Example 8-3You want to load a box into the back of a truck. One way is to lift it

straight up through a height h, as shown, doing a work W1. Alternatively, you can slide the box up a loading ramp a distance L, doing work W2. Assuming the ramp is frictionless, compare W1 & W2

Work done by gravity is independent of the path!

Negative Work

• If some component of the force (Fcosϕ) is in the opposite direction of the displacement, negative work will be done on the object– This occurs when 90° < ϕ ≤ 180

ϕd

F

Net Work (total Work)

• If more than one force acts on an object the net work Wnet (aka: total work) is the sum of the work done by each force separately

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Wnet =W1 +W2 +W3 +L = W∑

Example 8-4aA car of mass m coasts down a hill inclined at an angle θ below the

horizontal. The car is acted on by three forces: the normal force FN exerted by the road, the force due to air resistance Fair, and the force of gravity, mg. Find the total work done on it as it travels a distance d along the road

Net Work (total Work)• Alternatively, the net work can be calculated by

performing a vector sum of all the forces acting on the object to obtain the net force Fnet and then using our basic definition of work

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Wnet = Fnetdcosφ

Example 8-4bRepeat the calculation in Ex 7-4a using WNET = FNETd cosϕ

Kinetic Energy• Suppose we have an object accelerating from an

initial speed v0 to some new speed v through a distance d, they are related through the eqn:

• And by Newton’s 2nd Law we know:

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v 2 = v i2 + 2ad

or

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2ad = v 2 − v i2

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a =Fnet

m

Kinetic Energy• Combining the two eqns, we have:

• Multiplying each side by m/2 we have:

• And recalling that Fnetd = Wnet, we can write:€

2Fnet

m

⎛

⎝ ⎜

⎞

⎠ ⎟d = v 2 − v i

2( )

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Fnetd = 12mv

2 − 12mv i

2

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Wnet = 12mv

2 − 12mv i

2

Kinetic Energy• The quantity ½mv2 is called kinetic energy K, the

energy of motion

• Kinetic energy (like work) is a scalar • Depends on the speed v• Has units of Joules, J

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K = 12mv

2

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1 J = 1 kg ⋅m2 /s2

Work-Kinetic Energy Theorem• Taking K = ½mv2 and K0 = ½mv0

2 we can write:

• This is very useful and is known as the Work-Kinetic Energy Theorem

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Wnet = 12mv

2 − 12mv i

2

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Wnet =K f −K i

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Wnet = ΔK

Example 8-5A 4.10 kg box of books is lifted vertically from rest a distance of 1.60 m

with a constant, upward applied force of 52.7 N. Find the work done by the person lifting it, the work done by gravity, and the final speed of the box.

Example 8-6A boy exerts a force of 11.0 N at 29.0° above the horizontal on a 6.40-kg

sled. Find the work done by the boy on the sled. Find the final speed of the sled after it moves 2.00 m, assuming the sled starts with an initial speed of 0.500 m/s and slides horizontally without friction.

Graphical Interpretation of Work

Work = Area Under the Curve(of a F vs x graph)

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W = A

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W = bh

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W = Fd

Graphical Interpretation of Work for a Spring

Work = Area Under the Curve(of a F vs x graph)

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W = A

€

W = 12 bh

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W = 12 x kx( )

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W = 12 kx

2

Example 8-7Suppose a block has a mass of 1.5 kg and moves with an initial speed of

vi = 2.2 m/s. Find the compression of the spring, whose force constant is 475 N/m, when the block has momentarily come to rest.

Power

• Power is the time rate of work– How quickly energy is expended– Scalar quantity– Has SI units of Watts (W) or alt. horsepower (hp)

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1 watt = 1 W = 1 J/s = 1 kg ⋅m2 /s3

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1 horsepower = 1 hp = 746 W

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P =W

t

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P = FvOR

Table 7-3

Example 8-8To pass a slow-moving truck, you want your fancy 1.30×103-kg car to

accelerate from 13.4 m/s (30.0 mi/h) to 17.9 m/s (40.0 mi/h) in 3.00 s. What is the minimum power for this pass?

Example 8-10You are driving your 1300-kg car up a 15° incline at a constant speed

29.1 m/s. If the combined resistance (from friction and air resistance) is 650 N. Find the power output of your engine in horespower.