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HW6.1 – This is a lot of work!
WORK (make FBDs with the displacement vector shown clearly in order to calculate work)
1. A student holds her 1.5-kg psychology textbook out of a second floor classroom window until her arm is
tired; then she releases it.
a) How much work is done on the book by the student in simply holding it out the window?
No Work since there is no displacement of the book
b) How much work is done by the force of gravity during the time it falls 3.0 m?
WG  FG Dy cos 0
 mgDy cos 0  mgDy
 (1.5)(9.8)(3)  44.1J
d = 3m
Fg=mg
2. Joe horizontally pushes a 10-kg box at constant velocity of 3.0 m/s. The coefficient of kinetic
friction between the box and floor is 0.50.
a) How much work does Joe do if he pushes the
box for 15 meters?
WJoe  Fpush Dx cos 0
 f k Dx   k FN Dx   k mgDx
 (0.5)(10)(9.8)(15)
 735 J
b) How much work does friction do as Joe pushes the box?
Since Fnet = 0, Wnet = 0. Therefore
W fk  735 J
c) How far could Joe push the box at a constant velocity with an expenditure of 2000J?
Fpush  f k   k mg  49 N
WJoe  Fpush Dx cos 0
2000  49Dx
Dx  40.8m
3. Jane uses a vine wrapped around a pulley to lift a 70-kg Tarzan to a tree
house 9.0 meters above the ground.
a) How much work does Jane do when she lifts Tarzan?
Since Jane lifts Tarzan with constant velocity, Flift = Fg
Wlift  Flift Dy cos 0
Flift
 Fg Dy  mgDy  (70)(9.8)(9)
 6174 J
b)How much work does gravity do when Jane lifts Tarzan?
Since Fnet = 0, Wnet = 0. Therefore
Wg  Wlif t  6174 J
D y = 9m
Fg
4. A father pulls his child in a little red wagon with constant
speed. If the father pulls with a force of 16 N for 10.0 m,
and the handle of the wagon is inclined at an angle of 60o
above the horizontal, how much work does the father do
on the wagon? How much work does friction do on the
wagon? How much work does the normal force do on the
wagon? How much work does gravity do on the wagon?
W pull  Fpull Dx cos 60  Fpullx Dx
FN
Fpull y
fK
60o
 (16)(cos 60)(10)
 80 J
W fk  f k Dx cos180  W pull
 80 J
Fpull x
=FPcos60
Fg
WN  FN Dx cos 90  0
Wg  0
D x = 10m
FN
5. A bicycle rider pushes a bicycle that has a mass of 13 kg up a steep
hill. The incline is 250 and the road is 275 m long. The rider
pushes the bike parallel to the road with a force of 58 N.
a) How much work does the rider do on the bicycle?
FPush
W push  Fpush d cos 0
 (58)(275)
 15,950 J
25o
b) How much work is done by the force of gravity on
the bike?
3 ways to solve: THIS IS IMPORTANT TO UNDERSTAND
1. Definition of work: note that the angle between Fg
and d is (90+25) degrees (see right)
Fg
Wg  Fg d cos(90  25)  14,806 J
 mgd sin 25
(cos(90+) = -sin)
 mgh
h
25o
Fg//
sin25=h/d
2. get same expression if use only the component of Fg
parallel to d to calculate the work done by gravity:
Wg   Fg // d
25o
Fg
 (mg sin 25)(d )  14,806 J
 mgh
3. Get the same answer if you know that the work done by gravity is always mgDy regardless of the
path taken
Wg  mgh  14,806 J
WN  0
c) How much work is done by the normal force on the bike?
d) What is the net work on the bike?d
WNet  W push  WG  WN
 15950  14806  0  1144J
You would get the same answer by finding the work done by the net force (x-direction is along
incline):
WNet  Fnetx d  ( Fpush  Fgx )d
 (58  mg sin 25)(275)  1144J
The velocity-time graph of a particle of mass 2 kg moving in a
straight line is as shown in the graph at left. Find the work done
by all the forces acting on the particle (Wnet). (Don’t forget how
to analyze motion graphs!)
20
v (m/s)
6.
WNet   Fnet d  mad
0
0
t (s)
2
The net force is in the direction of the acceleration. Since the
object is slowing down in positive direction, the acceleration is
opposite the direction of displacement and velocity so angle
between Fnet and d is 180 and cos180=-1
There are 2 ways to proceed
a) Get a and d from the graph
a = slope of v-t = -10m/s2
d = area under v-t graph = ½(20)(2)=20m
WNet   Fnet d  mad  (2)(10 )(20 )  400 J
b) Use the W-E theorem
WNet  DK  K f  Ki
 12 mv 2f  12 mv02
 0  12 (2)(202 )  400 J
KINETIC ENERGY AND THE WORK ENERGY THEOREM
7. A 10.0 g bullet has a speed of 1200 m/s.
a) What is the kinetic energy of the bullet?
K  12 mv 2  12 (0.01)(1200 2 )  7200 J
b) How much work was done on the bullet to make it reach this speed if it started from rest?
Wnet  DK  12 mv f  12 mv0  7200 J
2
2
c) What is the bullet’s kinetic energy if the speed is halved? It is ¼ or 1800 J
d) What is the bullet’s kinetic energy if the speed is doubled? It is 4xs or 28,800 J
8. Two people, of equal weight, are running at 4 m/s and 5 m/s respectively. Both increase their speed by
1m/s in a time span of 10 s.
a) Who does more work?
Runner1: 4 to 5 m/s:
Wnet  DK  12 mv f  12 mv0
 12 m(52  42 )  4.5m
2
2
Runner2: 5 to 6 m/s: DOES MORE WORK
Wnet  DK  12 mv f  12 mv0
 12 m(62  52 )  5.5m
2
2
b) Who develops more power? Since both increase speed in same amount of time and P = W/t, Runner
2 develops more power.
c) How fast would each person have to go to double their kinetic energy? (How is v related to K?)
To double K, v would increase by a factor of 2 = 1.41.
2
K  12 mv
2K
v
m
Alternate solution:
Runner1 4 m/s
Runner 1 doubles K by increasing speed from 4 m/s to 5.64m/s
Runner 2 doubles K by increasing speed from 5 m/s to 7.05m/s
K  12 mv 2  8m
2 K  16 m
2
16 m  12 mv2
32  5.66  v2
d) By what factor does the runners’ kinetic energy increase if their speed is doubled? (How is K related
to v?) K increases 4 fold
9. A 1200-kg automobile travels at 25 m/s.
a. What is its kinetic energy?
K  12 mv 2  12 (1200 )25 2  375 ,000 J
b. What net work would be required to bring it to a stop?
Wnet  DK  K f  K i
 0  375,000  375000J
10. A 2.5 –g bullet traveling at 350 m/s hits a tree and slows uniformly to a stop while
penetrating a distance of 12 cm into the tree’s trunk. What force was exerted on the
bullet in bringing it to rest?
Wnet  DK  K f  K i
Ftree d cos180  mv  mv
1
2
2
f
1
2
2
0
 Ftree d  0  12 mv02
 Ftree (0.12)   12 (0.0025)(350 2 )
Ftree  1276 N
Ki
Ftree
Kf
d=0.12m
11. a) How much work is done by the force shown at right
when it acts on an object and pushes it from x = 0.25
m to x = 0.75 m? The force is not constant between
0.25m and 0.75 m so to get find the work done, must
take the area under the F-x curve
area = W = (0.4)(0.25) + (0.8)(0.25) = 0.3J
or
WF  Fav Dx  0.6(0.5)  0.3 J
b) Determine the final speed of a 1 kg object subjected
to this net force over this displacement if the initial
speed of the object at 0.25 m was 0.3 m/s.
Wnet  DK
0.3  12 mv f  12 mvi
2
2
 12 (1)v f  12 (1)0.320
2
v f  0.83m / s
12. How much work is done by the force shown below when it acts on an object and pushes it from x = 1.0
m to x = 4.0 m? The force is not constant between 1m and 4 m so to find the work done, must take the
area under the F-x curve
area = W1-4 = area of the shapes shown= 0.9J
Fx (N)
1.5
1
0.5
0
area = W0-8 = area of the shapes
shown= 5+10+2.5-2-4-2 = 9.5J
b) The stone starts from rest at x = 0. What
is its speed at x = 8 m (if the plotted
force is the net force)?
Fx (N)
0
13. A 2.0 kg stone moves along an x axis on a
horizontal frictionless surface, accelerated
by a force Fx(x) that varies with the stone’s
position as shown below
a) How much work is done on the stone by
the force as the stone moves from its
initial point at x1 = 0 to x2 = 8 m?
The force is not constant so to find the
work done, must take the area under the
F-x curve
5
4
3
2
1
0
-1 0
-2
-3
-4
1
10 J
5J
1
2
2
3
x (m)
3
4
2.5J
4
-2J 6
5
-4J
Wnet  DK
9.5  12 mv f  12 mvi
2
 12 (2)v f  0
2
9.5  3.1m / s  v f
2
x (m)
--2J
7
8