Ch. 5 – Work and Energy
(p.167-171) sec 5-1
(p.187-189) sec 5-4

A. Work
 transfer of energy through motion
 force exerted through a distance
W = Fd
W : work (J)
F : force (N) d : distance (m)
1 J = 1 N·m
Distance must be in direction of force!

A. Work
 Work is done when a force causes an object to move .
 Work is a scalar quantity.
 Work is the product of the force vector and displacement vector.
 Power is the rate at which work is done.

A. Work
Three things are necessary for the performance of work:
•
•
•
There must be an applied force, F .
There must be a displacement, Δx or d .
The force must have a component along the displacement .

A. Work
 Holding a 50 N bag of groceries is not work even though you get tired.
A force is exerted but the distance
(displacement) is zero.
 Carrying the same 50 N bag of groceries a distance of 10 m is not work either because there is no component of the weight vector that is parallel to the direction of motion.
work being done to overcome friction between your shoes and the floor

A. Work
When an object is being pulled across the floor by a rope held at an angle, the component of the force that is parallel to the direction of motion is
F•cos 
. d
θ
F
Fcos θ
In this case, work is being done because there is displacement in the direction of the force
W=Fcos
 •d

W = F•d [ ma Δx = (kg)•(m/s 2 )•(m)] the same variables are still with us!!!
 W work (Joules = kg•m 2 = N•m) s 2
 F magnitude of the force parallel to the direction of motion (Newton = kg•m) s 2
 d distance (m)

Example: A lawn mower is pushed a horizontal distance of 20 m by a force of
200 N directed at an angle of 30 0 with the ground. What is the work of this force?
F d = 20 m
30 0
Work = (F cos

) d
Work = (200 N) cos 30 0 (20 m)
Work = 3460 J
F = 200 N

Sample Problem
A man pulls a 50-kg stone a distance of 4 m at a constant velocity along a rough road by applying a 250-N force along a rope that makes an angle of 60
 with the horizontal. He then stoops down and lifts the stone vertically to a height of 1 m and carries it at a constant velocity a distance of 5 m. How much work does he do to the stone?

Solution


The man is doing work as he pulls the stone along the ground and again when he lifts the stone but no work is done when he is just carrying the stone.
To find the total work done in this problem you need to find the work done in each part then add them together.
W=F•d•cos  d = 4 m

= 60

F = 250 N
W = F•d = m•a•d m = 1 kg a = 9.8 m/s 2 d = 1 m

Solution
Answer: 990 J

B. Power
 work done per time
 force exerted through a distance
P = W/t
P : power (W)
W : work (J) t : time (s)
1 W = 1 J/s

B. Power
 Power is measured in Watts .
 Like work, power is a scalar quantity.
 Power is the rate at which work is done.
It is the work/time ratio. (P = W/t)
 For historical reasons, the horsepower is occasionally used to describe the power delivered by a machine.

1 horsepower = 745.699872 watts

B. Power
P

W t
P

Fd t
P

Fv
P – power (Watts = J/s)
W – work (Joules) t – time (seconds)
F – Force ( Newtons) d – distance (m) v – velocity (m/s) for the same amount of work, power and time are inversely proportional

Sample Problem
An elevator has a mass of 1000 kg and carries a maximum load of 800 kg. A constant frictional force of 4000 N retards its motion upward. What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed of 3 m/s?

m = 1800 kg v = 3 m/s
F = ma + friction
= (1800 kg)(9.8 m/s 2 ) + 4000N
= 21640 N
P = Fv
= 21640 N x 3 m/s
Answer: 64920 W

What power is consumed in lifting a 70-kg robber 1.6 m in 0.50 s?

What power is consumed in lifting a 70-kg robber 1.6 m in 0.50 s?
P

2
(70 kg)(9.8 m/s )(1.6 m)
0.50 s
Power Consumed: P = 2220 W

Solution
P = F v = ma g v
P = (900 kg)(9.8 m/s 2 )(4 m/s)
P = 35.3 kW

So far… mass = kg distance = Δx = m time = Δt = s velocity = Δx/Δt = m/s acceleration = Δv/Δt = m/s 2 force = m•a = Newton = kg•m/s 2 work = F•Δx = m•a•Δx = Joule = N•m = kg•m 2 /s 2 power = W/ Δt = F•Δx/Δt = F•v = Watt = J/s = kg•m 2 /s 3 notice how all of the units work with each other
1D and 2D equations can also be manipulated!

A. Work

Brett’s backpack weighs 30 N. How much work is done on the backpack when he lifts it 1.5 m from the floor to his back?
GIVEN:
F = 30 N d = 1.5 m
W = ?
WORK:
W = F·d
W = (30 N)(1.5 m)
W = 45 J
F
W d

A. Work
 A dancer lifts a 40 kg ballerina 1.4 m in the air and walks forward 2.2 m. How much work is done on the ballerina during and after the lift?
GIVEN: m = 40 kg d = 1.4 m - during d = 2.2 m - after
W = ?
F
W d
WORK:
W = F·d F = m·a
F =(40kg)(9.8m/s 2 )=392 N
W = (392 N)(1.4 m)
W = 549 J during lift
No work after lift. “d” is not in the direction of the force.

Sample Problem
A 65-kg athlete runs 600 m at a constant speed up a mountain inclined at 20
 with the horizontal. He performs this feat in 80 s. Assuming that air resistance is negligible, (a) how much work does he perform, and (b) what is his power output during the run?
W=F•d P =W/t

Sample Problem
A 65-kg athlete runs 600 m at a constant speed up a mountain inclined at 20
 with the horizontal. He performs this feat in 80 s. Assuming that air resistance is negligible, (a) how much work does he perform, and (b) what is his power output during the run?
W=F•d
W = ma y
•d y
(hint) d=600 m
20 °
F = ma since F is in y direction, need distance in y direction

Answer: 130720 J, 1634 W

Sample Problem
A 20000 kg truck goes up a 20 o incline of 500 m in
1 minute. How much power (in Watts) was shown by the truck?

Answer: 558632 Watts

B. Power
 A person is also a machine which has a power rating . Some people are more power-full than others.
 That is, some people are capable of doing the same amount of work in less time or more work in the same amount of time.