Download Homework #1: Work

Homework #1: Work {Required Problems: All of the first 6 problems.}
1.
A weight lifter lifts a 350-N set of weights from ground level to a position
over his head, a vertical distance of 2.00 m. How much work does the weight
lifter do, assuming he moves the weights at constant speed?
2.
If a man lifts a 20.0-kg bucket from a well and does 6.00 kJ of work, how
deep is the well? Assume that the speed of the bucket remains constant as it is
lifted.
3.
A tugboat exerts a constant force of 5.00 × 103 N on a ship moving at
constant speed through a harbor. How much work does the tugboat do on the
ship if each moves a distance of 3.00 km?
4.
(I) How much work is done by the gravitational force when a 265-kg pile driver falls
2.80 m?
5.
(I) A 1300-N crate rests on the floor. How much work is required to move it at
constant speed (a) 4.0 m along the floor against a friction force of 230 N, and (b) 4.0
m vertically?
6.
(II) A box of mass 5.0 kg is accelerated by a force across a floor at a rate of 2.0 m s 2
for 7.0 s. Find the net work done on the box.
Extra Problems (not required)
5.
Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0°
incline. The coefficient of kinetic friction between the block and the incline is μk =
0.436. Determine (a) the work done by the force of gravity, (b) the work done by
the friction force between block and incline, and (c) the work done by the normal
force.
6.
A horizontal force of 150 N is used to push a 40.0-kg packing crate a
distance of 6.00 m on a rough horizontal surface. If the crate moves at constant
speed, find (a) the work done by the 150-N force and (b) the coefficient of kinetic
friction between the crate and surface.
7.
A sledge loaded with bricks has a total mass of 18.0 kg and is pulled at
constant speed by a rope inclined at 20.0° above the horizontal. The sledge
moves a distance of 20.0 m on a horizontal surface. The coefficient of kinetic
friction between the sledge and surface is 0.500. (a) What is the tension in the
rope? (b) How much work is done by the rope on the sledge? (c) What is the
mechanical energy lost due to friction?
7.
(II) A lever such as that shown in Fig. 6–35 can be used to lift objects we might not otherwise be able
to lift. Show that the ratio of output force, FO , to input force, FI , is related to the lengths l I and l O
from the pivot point by FO FI  l I l O (ignoring friction and the mass of the lever), given that the
work output equals work input.
9.
(II) (a) Find the force required to give a helicopter of mass M an acceleration of 0.10 g upward. (b)
Find the work done by this force as the helicopter moves a distance h upward.
10. (II) What is the minimum work needed to push a 950-kg car 810 m up along a 9.0º incline? (a) Ignore
friction. (b) Assume the effective coefficient of friction retarding the car is 0.25.
Figure 6-35
Solutions #1 Work
1
If the weights are to move at constant velocity, the net force on them must be
zero. Thus, the force exerted on the weights is upward, parallel to the
displacement, with magnitude 350 N. The work done by this force is
W   F cos  s  350 N  cos0  2.00 m   700 J
2
To lift the bucket at constant speed, the woman exerts an upward force whose
magnitude is F  m g   20.0 kg 9.80 m s2  196 N . The work done is


W   F cos  s, so the displacement is
s
W
6.00  103 J

 30.6 m
F cos 196 N  cos0




3
W   F cos  s  5.00  103 N cos0 3.00  103 m  1.50  107 J 15.0 M J
4.
The force and the displacement are both downwards, so the angle between them is 0o.

WG  mgd cos    265 kg  9.80 m s 2
5.
  2.80 m  cos 0
o
 7.27 103 J
(a) See the free-body diagram for the crate as it is being pulled. Since the
crate is not accelerating horizontally, FP  Ffr  230 N . The work done to
move it across the floor is the work done by the pulling force. The angle
between the pulling force and the direction of motion is 0o.
x
Ffr
WP  FP d cos 0o   230 N  4.0 m 1  9.2  10 2 J
WP  FP d cos 0o  mgd  1300 N  4.0 m   5.2  103 J
Since the acceleration of the box is constant, use Eq. 2-11b to find the distance moved.
Assume that the box starts from rest.

x  x  x0  v0t  12 at 2  0  12 2.0 m s 2
 7 s
Then the work done in moving the crate is

W  F x cos 0o  max   5 kg  2.0 m s 2
2
mg
FN
(b) See the free-body diagram for the crate as it is being lifted. Since the crate is not
accelerating vertically, the pulling force is the same magnitude as the weight. The
angle between the pulling force and the direction of motion is 0o.
6.
FP
 49 m
  49 m   4.9 10 J
2
y
FP
mg
5.5


(a) The force of gravity is given by m g   5.00 kg 9.80 m s2  49.0 N and is
directed downwards. The angle between the force of gravity and the
direction of motion is  = 90.0° - 30.0° = 60.0°, and so the work done by
gravity is given as
W g   F cos  s  49.0 N  cos60.0  2.50 m   61.3 J
(b) The normal force exerted on the block by the incline is n  m g cos30.0 , so
the friction force is
fk  kn   0.436 49.0 N  cos30.0  18.5 N
This force is directed opposite to the displacement (that is  = 180°), and the
work it does is
W f   fk cos  s 18.5 N  cos180  2.50 m    46.3 J
(c) Since the normal force is perpendicular to the displacement;
  90, cos  0 , and the work done by the normal force is zero .
5.7
(a)
Fy  F sin  n  m g  0
n  m g  F sin 
Fx  F cos  kn  0
n

F cos
k
m g  F sin 
F cos
k
 0.500 18.0 kg  9.80 m s 
km g
=
= 79.4 N
k sin  cos  0.500 sin 20.0  cos20.0
2
F=
(b) W F   F cos  s  79.4 N  cos20.0  20.0 m   1.49  103 J 1.49 kJ
fk  F cos  74.6 N
(c)
W
f
=  fk cos  s= 74.6 N  cos180  20.0 m   1.49  103 J  1.49 kJ
7.
Consider the diagram shown. If we assume that the man pushes
straight down on the end of the lever, then the work done by the
man (the “input” work) is given by WI  FI hI . The object moves a
FI
shorter distance, as seen from the diagram, and so WO  FO hO .
Equate the two amounts of work.
FO hI
FO hO  FI hI 

FI hO
But by similar triangles, we see that
9.
(a)
hI
hO

lI
lO
, and so
FO
FI

lI
hI
lO
lI
lO
hO
FO
.
Write Newton’s 2nd law for the vertical direction, with up as positive.
 Fy  FL  Mg  Ma  M  0.10 g   FL  1.10 Mg
FL
(b) The lifting force and the displacement are in the same direction, so the work
done by the lifting force in lifting the helicopter a vertical distance h is
Mg
WL  FL h cos 0o  1.10 Mgh .
10. Draw a free-body diagram of the car on the incline. Include a frictional
force, but ignore it in part (a) of the problem. The minimum work will
occur when the car is moved at a constant velocity.
(a) Write Newton’s 2nd law in both the x and y directions, noting that the
car is unaccelerated.
 Fy  FN  mg cos  0  FN  mg cos 
F
x
y
FN
FP
Ffr
 FP  mg sin   0  FP  mg sin 


mg
The work done by FP in moving the car a distance d along the plane (parallel to FP ) is
given by

WP  FP d cos 0o  mgd sin    950 kg  9.80 m s 2
 810 m  sin 9.0
o
 1.2 106 J
(b) Now include the frictional force, given by Ffr   k FN . We still assume that the car is
not accelerated. We again write Newton’s 2nd law for each direction. The y-forces are
unchanged by the addition of friction, and so we still have FN  mg cos  .
F
x
 FP  Ffr  mg sin   0  FP  Ffr  mg sin   k mg cos  mg sin  .
The work done by FP in moving the car a distance d along the plane (parallel to FP ) is
given by
WP  FP d cos 0o  mgd  sin    k cos  

  950 kg  9.80 m s 2
 810 m  sin 9.0
o

 0.25cos 9.0o  3.0  106 J
x