Transcript I. Work - District 196

Ch. 5 – Work and Energy
I. Work (p.167-171) sec 5-1
II. Power (p.187-189) sec 5-4
A. Work
 transfer of energy through motion
 force exerted through a distance
W = Fd
W:
F:
d:
work (J)
force (N)
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
F
θ
Fcosθ
In this case,
work is being
done because
there is
displacement in
the direction of
the force
W=Fcos•d
A. Work
W = F•d
[maΔx = (kg)•(m/s2)•(m)]
the same variables are still with us!!!
W
F
d
work (Joules = kg•m2 = N•m)
s2
magnitude of the force
parallel to the direction of
motion (Newton = kg•m)
s2
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 300 with the
ground. What is the work of this force?
F
d = 20 m
300
Work = (F cos  ) d
Work = (200 N) cos 300 (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
W = F•d = m•a•d
d=4m
 = 60
F = 250 N
m = 1 kg
a = 9.8 m/s2
d=1m
Solution
Answer: 990 J
B. Power
 work done per time
 force exerted through a distance
P = W/t
P:
W:
t:
power (W)
work (J)
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 746 Watts.

1 horsepower = 745.699872 watts
B. Power
W
P
t
Fd
P
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
P = Fv
= 21640 N x 3 m/s
F = ma + friction
= (1800 kg)(9.8 m/s2) + 4000N
= 21640 N
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 = F•d/Δv
P = ma•d/Δt
2
(70 kg)(9.8 m/s )(1.6 m)
P
0.50 s
Power Consumed: P = 2220 W
What power is required to lift a 900-kg
elevator at a constant speed of 4 m/s?
Solution
What power is required to lift a 900-kg
elevator at a constant speed of 4 m/s?
P = F v = magv
P = (900 kg)(9.8 m/s2)(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/s2
force = m•a = Newton = kg•m/s2
work = F•Δx = m•a•Δx = Joule = N•m = kg•m2/s2
power = W/Δt = F•Δx/Δt = F•v = Watt = J/s = kg•m2/s3
notice how all of the units work with each other
1D and 2D equations can also be manipulated!
Following are a few
more examples you
may want to try.
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
W
F 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=?
W
F d
WORK:
W = F·d
F = m·a
F =(40kg)(9.8m/s2)=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 = may•dy
d=600 m
20°
(hint)
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 20o 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.
