Work = Force X distanceW = Fd

W = Fdcos

• Unit – Joules• Force must be direction of motion

WNET = KE

Work or Not

1. A teacher pushes against a wall until he is exhausted.

2. A book falls off the table and falls freely to the ground.

3. A waiter carried a full try of meals across the room.

4. A rocket accelerates through space.

Mr. Fredericks pulls a 10 kg box with 30 N of Force a distance of 50 m, at an angle of 50o with the ground.

a. Calculate the work that was done (964 J)

b.Calculate the normal force on the suitcase. (75 N)

= 50o

Direction of motion

Work: Example 4

A 50-kg crate is pulled 40 m with a force of 100 N at an angle of 37o. The floor is rough and exerts a frictional force of 50 N. Determine the work done on the crate by each force and the net work done on the crate.

Fp

Ffr FN

mg

A 150,000 kg rocket launches straight up with a thrust of 4.0 X 106 N.

a. Calculate the work done by thrust at 500 m. (2.0 X 109 J)

b.Calculate the work done by gravity. (-7.4 X 108 J)

c. Calculate the net work. (1.26 X 109 J)d.Calculate the speed of the rocket. (130 m/s)

A 500 g air hockey puck slides across an air table at 2.0 m/s. The player blows on the puck at an angle of 30o to the horizontal with a force of 1.0 N for 50 cm. The player is trying to slow the puck.

a. Calculate the work done by the player. (-0.433 J)b.Calculate the final speed of the puck (1.5 m/s)

Work: Variable Force

Work is really an area:

W =∫Fdx

(an integral tells you the area)

WORK

The magnitude of a force on a spring varies according to F(x) = 1500x2. Calculate the work done stretching the spring 10 cm from its equilibrium length.

W =∫00.10m Fdx

W =∫00.10m 1500x2dx

W = 500x3| 00.10m = 0.50 Joules

A 1500 kg car accelerates from rest. The graph below shows the force on the car.

a. Calculate the work done on the car. (1 X 106 J)b.Calculate the speed after 200 m. (37 m/s)

A 100 g pinball is launched by pulling back a 20 N/m spring a distance of 20 cm. However, there is friction and k = 0.10.

a. Calculate the work done by the spring. (0.400 J)b.Calculate the work done by friction. (-0.020 J)c. Calculate the speed of the ball on release. (2.8

m/s)

Does the Earth Do Work on the Moon?

W = FdcosW = Fd(cos 90o)W = Fd(0)W = 0

FR

v

English Unit of Work

• Foot-pound – English unit of work.• Pound – unit of Force• Foot – Unit of distance• W = Fd = (foot*pound)

A 70 kg is gliding at 2.0 m/s when he starts down a slippery 10o slope. He travels for 50 m.

a. Calculate the force parallel to the ground pulling him down the hill.

b.Calculate work done by gravity for the 50 m.c. Calculate his speed at the bottom. Remember

that initially he was not at rest.

Conservative and Nonconservative Forces

Conservative Forces– Work is independent of the path taken– Gravity, electromagnetic forces

Nonconservative Forces– Work depends on the path taken– Friction (dissipative forces)

Nonconservative ForcesWill it take more work to push the box on path A

or path B? Or are they the same?

B

A

If nonconservative forces act, use:

KE1 + PE1 = KE2 + PE1 + Wfr

½ mv2 + mgy = ½ mv2 + mgy + Ffrd

Mr. Fredericks (100 kg) slides down a 3.5 m tall slide. If he leaves the slide at the bottom at 6.3 m/s, what is the Force of friction and the coefficient of friction for the slide? Assume the slide is 6.0 m long. (0.25)

3.5 m 6.0 m

A 70 kg skier starts at the top of the slope at 2.0 m/s. The slope is 50 m long and has an elevation of 10o. There is a wind exerting a 50 N retarding force at the bottom.

a. Calculate the work done by gravity b.Calculate the work done by the retarding forcec. Calculate his speed at the bottom (10 m/s)

A 5.0 kg box is attached to one end of a spring (80 N/m). The other end is attached to the wall. The spring is stretched 50 cm by a constant force of 100 N. There is friction and k = 0.30.

a. Calculate the work done by the pullb.Calculate the work done on the springc. Calculate the work lost to friction (thermal

energy)d.Calculate the speed of the box at 50 cm (3.6

m/s)

Force and Potential Energy

F = - dU ds• Force is the negative of the derivative of the

potential energy.• Force is the negative slope.

Example:Calculate the gravitational force for gravitational

Potential energy (mgy)

Calculate the force being exerted on a particle given the following potential energy curve:

Given the following potential energy graph, sketch the force versus distance graph.

Power

Power = Work P = W time t

• Metric Unit: Joules/s = Watt.• Definition – rate at which work is done– A powerful engine can do a lot of work quickly.– Running and walking up the steps require the same

amount of work.– Running up steps requires more Power

a. A donkey performs 15,000 J of work pulling a wagon for 20 s. What is the donkey’s power?

b. What power motor is needed to lift a 2000 kg elevator at a constant 3.0 m/s? (Hint: use 1 second in your calculations)

c. A motor and cable drags a 300 kg box across a rough floor at 0.50 m/s. The coefficient of kinetic friction is 0.60. Calculate the necessary power.

Horsepower

• The English Unit of power is horsepower

• Foot-lb = Horsepower (hp)second

• 1 hp = 746 Watts• 1 hp = ½ Columbus (who sailed in 1492)

1. How much horsepower is required to power a 100 Watt lightbulb?

2. A 1500 kg car has a profile that is 1.6 m wide and 1.4 m high. The coefficient of rolling friction is 0.02.

1. Calculate the drag force if the car travels at a steady 30 m/s (1/4Av2) (504 N)

2. Calculate the force the car must exert against drag and friction. (798 N)

3. Calculate the power the engine must provide if 25% of the power is lost between the engine and the wheels.

Horsepower

Consider a 40 hp car engine that can go from 0 to 60 mi/hr in 20 seconds.

A 160 hp car could go from zero to 60 mi/hr in 5 seconds.

4 times as powerful means it can do the same work in ¼ the time.

Horsepower: Example 4

A crane lifts a 200 N box 5 meters in 3 seconds. What is the crane’s power in Watts and in horsepower?

Power and Calculus

P = W tWork = FdP = Fd tP = Fv

Power and Calculus: Ex 1

Find the power delivered by a net force at t=2 s to a 0.5 kg mass that moves according to x(t) = 1/3t3

v = dx/dt v = t2 v = (2)2 = 4 m/sa = dv/dta = 2t a = 4 m/s2

F = maF = (0.5 kg)(4 m/s2) = 2 N

P = Fv = (2N)(4 m/s)

Springs and Calculus

The force in a spring is variable (F = -kx)Work = ∫0

x F(x) dxWork = ∫0

x -kx dxWork = - ½kx2

Work = -PEPE = ½ kx2