Work and Energy
Definition of Work (W)
Work is done when a force is exerted on an
object AND the object moves in the direction
of the force (or one of its components)!
Is Work done (to the object) when...
•when you push box across the floor?
YES- applied force in direction of distance
covered
YES- F is up
•when you lift a briefcase? and so is ∆d
•by the normal force acting on a sliding block?
NO- F is up, ∆d is across!
The amount of work done depends directly upon
the amount of force that is doing the work and
how far the object is moved:
units of work:
W = F∆d or Fcosø(∆d)
J = Nm
A Joule (J) is the standard unit of work (as well
as the standard unit of energy) and it equals
Newtons times meters!
• Work is the scalar product of two vectors!
• Work has no direction, BUT work done
opposite the motion is considered negative!
When determining work, you need to consider
which force:
Fa does (+) work
N
Ff
Fa
Fp
ø
Fw
ø Fn
Gravity and friction
do negative work:
How much work is
done by the normal
force?
0 J N is 90˚ from ∆d
The amount of work (energy expended) done
in each case is the same– less force is used
over a greater distance, but ultimately the
same work is done!
Calculate how much work is done by the applied
force when a crate weighing 850.0 N is pushed up
a 15.0˚ incline at constant speed for a distance of
10.0 m. The µ between the crate and the plane is
.240.
N
Fa
W=?
Fp
Fw = 850.0 N
Ff
ø = 15.0˚
ø
∆d = 10.0 m
Fn
ø
µ = .240
Fw
Balanced Forces?
Fa = Ff + Fp
W = Fa∆d
Yes- constant speed!
N = Fn
Fp = sinø(Fw) = sin(15.0˚)(850 N)
220 N
Fn = cosø(Fw)= cos15.0˚(850 N)
= 821 N
Ff = µFn = .240(821 N) = 197 N
Fa = Ff + Fp = 197 N + 220 N
= 417 N
= (417 N)(10.0 m)
= 4170 J
A girl pulls a wagon with constant velocity along a
level path for a distance of 45.0 m. She pulls with
85.0 N of force along the handle which makes an
angle of 20.0˚ with the horizontal. How much
work does she do?
N
∆d = 45.0 m
Fa
Fa = 85.0 N
Ff
Fv
ø
ø = 20.0˚
Fh
W=?
Fw
Fh = Ff
N + Fv = Fw
W = Fh∆d
Fh = cosø(Fa)
= cos(20.0˚)(85.0 N)
= 79.9 N
= (79.9 N)(45.0 m)
= 3600 J
1) A 25.0 kg box slides at constant speed down a
ramp of 15.0˚. How much work is done by
gravity if the ramp is 8.00 m long?
507 J
2) A car traveling 20.0 m/s slams on the brakes
and stops in a distance of 12.0 m. If the car
weighs 14,100 N, how much work is done by
friction in stopping the car? -288,000 J
3) A 245 N crate is pulled by a rope up a 20.0˚
incline at constant speed. How much work is
done by the rope if the crate is pulled up 6.00 m
and µ = .250.
848 J
4) A crate that weighs 250.0 N accelerates from
rest down a 20.0˚ incline until it reaches 5.00 m/s
at the bottom. How much work was done to the
708 J
box by gravity if µ = .200?
5) A 200.0 N sled is accelerated from rest to 15.0
m/s in 20.0 s. If the rope used to pull the sled is
tied horizontally, how much work is done by the
rope to the sled during this 20.0 s if µ = .225 ?
9050 J
6) A man accelerates a 25.0 kg crate from rest up
a 2.00 m ramp in 5.00 s. How much work does
the man do if the ramp is at an incline of 15.0˚
with horizontal and µ = .250?
253 J
Fat guy: 300 lbs
Skinny guy: 150 lbs
Who does more work
going up these stairs if the
fat guy takes 10 s and the
skinny guy takes 5 s?
Power (P)
•does not refer to strength as is the common
definition!
•the timed rate at which work is done (or
the rate at which energy is expended)
•depends directly upon the work and
inversely upon the time to do that work!
P=
W
∆t
=
F∆d
∆t
units for power:
J
s
=w
one watt (w) equals one
Joule (J) per second (s)
Power can also be measured in Horsepower
(Hp).
1 Hp = 746 w
(25 w)(1 Hp/746 w) = .0335 Hp
A 35.0 kg box falls off of a truck that is traveling at
25.0 m/s and slides to a stop over a distance of 140
m. How much power is generated by friction
during this time?
N
m = 35.0 kg
Ff
Fw = 343 N
v0 = 25.0 m/s
v=0
∆d = 140 m
P=?
Fw
Ff = Fnet
Fw = N
P = Ff ∆d
∆t
Ff = ma
= (35.0 kg)(-2.23 m/s2)
= -78.1 N
= (-78.1 N)(140 m)
11.2 s
= - 976 w
negative power means
the work done is
opposite of the motion!
a = vf2 -vi2
2∆d
= 0 - (25.0 m/s)2
2(140 m)
= -2.23 m/s2
∆t = vf - vi
a
= 0 - 25.0 m/s
-2.23 m/s2
= 11.2 s
A 45.0 kg crate is accelerated from rest up a 20.0˚
incline at 2.00 m/s2. At what rate was the work
done by the applied force if the ramp is 6.00 m
long and it took 7.00 s to reach the top (µ = .230)?
N
Fa
m = 45.0 kg
vi = 0
Fp
Ff
ø = 20.0˚
a = 2.00 m/s2
P=?
∆d = 6.00 m
Fn
∆t = 7.00 s
Fw
µ = .230
Fa > Fp + Ff
N = Fn
Fa = Fp + Ff + F
P = Fa∆d
∆t
= 288 w
Fp = sinø(Fw)
= sin(20.0)(441 N) = 151 N
N = Fn = cosø(Fw)
= cos(20.0)(441) = 414 N
Ff = µN = (.230)(414 N) = 95.3 N
Fnet = ma = 45.0 kg(2.00 m/s2)
= 90.0 N
Fa = 151 N + 95.3 N + 90.0 N = 336 N
1) A 4.00 m ramp is used to pull a 25.0 kg box up
a 15.0˚ incline at constant speed. At what rate
was the work done if it took 5.00 s to get the
crate to the top and µ = .230?
2) A 44.0 kg crate is allowed to slide from rest
down a 3.50 m ramp until it reaches the bottom of
the 18.0˚ incline 2.64 s later. At what rate was the
work done if µ = .200?
3) A 45.0 kg crate is pulled across a level floor (µ
= .220) by a force of 155 N applied to a rope that
is tied at a 25.0˚ to the horizontal. The crate is
pulled from rest for 6.00 s. How much power is
generated during this time?
4) A 452 N crate is pulled across a level floor at
constant speed by a force of 135 N applied to a
rope tied at a 30.0˚ angle to the horizontal. How
much power is generated if the crate is pulled a
distance of 12.0 m in 15.0 s?
5) How much work is done in accelerating a 35.0
kg crate at 2.00 m/s2 up a 16.0˚ incline if the
incline is 5.00 m long and µ = .210?
6) A 34.5 kg crate falls off the back of a truck
traveling 18.0 m/s and slides to a stop in 7.00 s.
What is the average power generated by friction
during this time?
Energy
• general term that refers to the ability
to do work!
• there are MANY ways to classify
types of energy!
• the two most fundamental are:
potential energy
kinetic energy
Potential Energy
•stored up energy- the potential to do
some kind of work
•that potential can come from a variety of
sources:
gravitational PEpotential due to position and
gravity
elastic PEpotential due to a stretched or
compressed elastic object
chemical PEpotential due to chemical reaction
nuclear PEpotential due to nuclear chain reaction
Gravitational Potential Energy (Ep)
• potential energy because gravity could
move an object a certain distance
• depends upon mass, gravity, and height
• height is determined from a set reference
point- and that point can change!
units are
Ep = mg∆h = Fw∆h
Nm = J
the standard unit of energy is the Joule (J)!
How much potential energy would a 2.0 kg book
have if it were held 1.0 m above a desktop that is
1.0 m above the floor that is 4.0 m above the
ground?
Ep = mg∆h
relative to the desk:
Ep = (2.0 kg)(9.80 m/s2)(1.0 m)= 20 J
relative to the floor:
Ep = (2.0 kg)(9.80 m/s2)(2.0 m) = 40 J
relative to the ground:
Ep = (2.0 kg)(9.80 m/s2)(6.0 m) = 120 J
Unless specifically told otherwise, gravitational
Ep is relative to the surface of the earth!
Elastic Potential Energy (Ep)
• potential energy due to a stretched or
compressed elastic object - like a spring!
• in order to stretch or compress a spring, the
force applied must increase with the amount of
stretch/compression of the spring.
2N
1 cm
4N
2 cm
3 cm
6N
4.5 cm
9N
the relationship here is constant!
The ratio of the amount of force needed to stretch
(or compress) a spring a certain amount is called
the force constant! (k)  Hooke’s Law
k=
F
units 
∆d
N
m
If 50.0 N is used to stretch a spring a total
distance of 10.0 cm:
k=
50.0 N =
5.00 N = 25.0 N
.100 m
.0100 m
.0500 m
= 500 N/m
If 40.0 N is used to stretch a spring 10.0 cm,
what is the force constant?
k=F
= 40.0 N
∆d
.100 m
= 400 N/m
How much work was done in stretching the
spring?
W = F∆d = (40.0 N)(.100 m) = 4.00 J
WRONG!
The 40.0 N of force was the final force reached, it
increased steadily from 0 N.
The average force applied then was :
Fav = Fi + Ff
2
= 0 + 40.0 N
2
= 20.0 N
With a spring, work is calculated using the
average force (because the force constantly
changes):
W = Fav∆d = (20.0 N)(.100 m) = 2.00 J
The work done in stretching the spring will equal
the gain in elastic potential energy. Therefore the
spring will gain 2 J of potential energy:
So how do you find the PE of a spring?
Ep = .5k∆d2
• for the given spring:
k=F
= 40 N
∆d
.1 m
= 400 N/m
units  J
Ep= .5k∆d2
= .5(400 N/m)(.1 m)2
=2J
A weight of 5.5 N will compress a spring exactly
1.0 cm. How much potential energy will the
spring have when it is compressed 4.2 cm?
F = 5.5 N
∆x1 = .010 m
∆x2 = .042 m
Ep = ?
k=F
∆x1
Ep = .5k∆x22
.5(550 N/m)(.042 m)2
.49 J
= 5.5 N
.010 m
= 550 N/m
If an object is moving for any reason, then it is
said to have kinetic energy.
Kinetic energy (Ek) is directly related to the mass
of the object as well as the square of the velocity!
Ek = .5mv2
Law of Conservation of Energy
In a closed system (no energy lost to forces
such as friction), the total amount of energy
(the KE and PE) must remain constant!
Ep + Ek = Et
or
∆Ep + ∆Ek = 0
Energy is never lost- only converted from one
form to another!
W = ∆K -> The WorkEnergy Theorem
A rock that weighs 1120 N is at the edge of a cliff
that is 22.0 m high. How much Ep and Ek does the
rock have at the top of the cliff?
Fw = 1120 N
Ep = mg∆h = (1120 N)(22.0 m)
m = 114 kg
= 24, 600 J
∆h = 22.0 m
Ek = 0 J (it is not moving!)
v=0
For this system, the total energy (Et):
Et = Ep + Ek = 24,600 J + 0 J = 24,600 J
How much Ep and Ek does it have as it hits after
falling to the ground?
h=0
Ep = mgh = 0J
Ek = 24,600 J
Et = Ep + Ek = 0 + 24,600 J = 24,600 J
A check of this answer:
vi = 0
∆d = -22.0 m
a = -9.80 m/s2
Ek = ?
Ek = .5mv2
= .5(114 kg)(20.8 m/s)2
= 24,600 J
vf = vi2 + 2a∆d
= 0 + 2(-9.80 m/s2)(-22.0 m) = 20.8 m/s
What speed will the object have when it has fallen
half way?
Solving using conservation of energy:
Ek = .5 mv2
Et = 24,600 J
∆h = 11.0 m
Ep = 12, 300 J
Ek = 12, 300 J
v=?
v=
=
Ek
.5m
12,300 J
.5(114 kg)
= 14.7 m/s
Checking as a freefall problem:
vi = 0
vf = vi2 + 2a∆d
∆d = -11.0 m
2)(-11.0 m)
=
2
(-9.80
m/s
a = -9.80 m/s2
vf = ?
= 14.7 m/s
What speed will the object have after it has fallen
a distance of 6.7 m?
∆h = 22.0 - 6.7 m = 15.3 m
Ep = mg∆h = (1120 N)(15.3 m) = 17,100 J
Ek = Et - Ep = 24, 600 J - 17,100 J = 7500 J
v=
Ek
.5m
=
7500 J
= 11.5 m/s
.5(114 kg)
A spring with a force constant of 1560 N/m is
compressed 12.0 cm and used to launch a mass of
1.00 kg horizontally. With what speed will the
mass leave the spring?
k = 1560 N/m
∆x = .120 m
m = 1.00 kg
Fw= 9.80 N
v=?
The Ep of the compressed spring will be turned
into Ek of the launched object!
Ep = .5k∆d2
Ek = .5mv2
= .5(1560 N/m)(.120 m)2
= 11.2 J
Ek
v=
.5m
=
11.2 J
= 4.73 m/s
.5(1.00 kg)
How high above the vertically compressed
spring will the object be launched?
Ans: 1.14 m
Solve using the conservation of energy:
A) A spring is used to launch a 2.50 kg
projectile with a horizontal speed of 15.0 m/s.
If the spring is compressed 30.0 cm before the
launch, what is the force constant of the spring?
6250 N/m
B) A 125 kg rock is dropped from a cliff so that
it hits the ground with a speed of 42.0 m/s.
How high is the cliff from which it was
dropped?
90.0 m
1) A bowstring of force constant 1230 N/m is
stretched 30.0 cm. What is the weight of an
arrow that is launched horizontally with a velocity
of 25.0 m/s? (vertically?)
2) A boulder of mass 255 kg is sitting atop a hill
that is 34.6 m high. The boulder rolls down that
hill and over the next hill, which is only 18.9 m
high. Assuming the boulder loses no energy to
friction and such, what is the speed of the boulder
as it passes over the next hill?
3) A spring of constant 1250 N/m is compressed
45.0 cm and used to launch a 5.00 kg box up in the
air. If the coefficient of friction is .250, how far
above the uncompressed spring will the box rise
before stopping?
4) A 42.0 kg box slides from rest down a
frictionless 25.0˚ incline. It accelerates for 3.48 s
to reach the bottom of the incline where it
encounters a spring bumper that is compressed
40.0 cm in stopping the box. What is the force
constant of this spring?
5) A spring of force constant 655 N/m is used to
launch a 5.25 kg mass from the edge of a
frictionless table that is 1.20 m above the floor.
How far horizontally away from the table will the
mass land if the spring was compressed 42.0 cm
before launching?
2.32 m