5.9
(a) The work-energy theorem, Wnet = KE f − KEi , gives
5000 J =
(
)
1
2.50 × 103 kg v 2 − 0 , or
2
v = 2.00 m s .
(b) W = ( F cos θ ) s = ( F cos 0°) ( 25.0 m ) = 5000 J , so F = 200 N .
A 70-kg base runner begins his slide into second base when moving at a speed of 4.0 m/s.
The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is
zero just as he reaches the base.
(a) How much mechanical energy is lost due to friction acting on the
runner?
Enter your answer in joules without the units in scientific notation
rounded to two significant figures.
(b) How far does he slide?
Enter your answer in meters without the units rounded to two significant
figures. Take the gravitational acceleration to be
5.13
(a) We use the work-energy theorem to find the work.
1
1
1
2
W = ΔKE = mv 2 − mv i2 = 0 − ( 70 kg )( 4.0 m s ) = − 5.6 × 102 J .
2
2
2
(b) W = ( F cos θ ) s = ( fk cos180°) s = ( − μk n ) s = ( − μk mg ) s ,
so
(
)
−5.6 × 102 J
W
s=−
=−
= 1.2 m .
μk mg
( 0.70) ( 70 kg ) 9.80 m s 2
(
)
A bead of mass m = 4.70 kg is released from point A and slides on the frictionless track shown in the
following figure.
(a) Determine the bead's speed at points B and C.
Round your answers to three significant figures. Take the free fall acceleration to be 9.80
.
m/s, point C:
m/s
point B:
(b) Determine the net work done by the force of gravity in moving the bead from A to C.
Round your answer to three significant figures.
J
5.30
(a) From conservation of mechanical energy,
1 2
1
mv B + mgy B = mv A2 + mgy A , or
2
2
A
B
5.00 m
v B = v A2 + 2 g ( y A − y B )
(
)
= 0 + 2 9.80 m s 2 (1.80 m ) = 5.94 m s .
Similarly,
v C = v A2 + 2 g ( y A − y B ) = 0 + 2 g ( 5.00 m − 2.00 m ) = 7.67 m s .
(b) W g
)
A →C
( ) − ( PE )
= PEg
A
g
C
= mg ( y A − y C ) = ( 49.0 N )( 3.00 m ) = 147 J
3.20 m
C
2.00 m
A 670-kg elevator starts from rest. It moves upward for 2.85 s with constant acceleration until it reaches its
cruising speed, 2.20 m/s.
(a) What is the average power of the elevator motor during this interval?
(Hint: 1 hp = 746 W.)
Round your answer to three significant figures. Take the gravitational acceleration to be
hp
(b) How does this compare with its power during an upward cruise with constant speed?
Round your answer to three significant figures.
hp
5.54
(a) The acceleration of the elevator during the first 3.00 s is
a=
v f − vi
t
=
1.75 m s − 0
= 0.583 m s 2 ,
3.00 s
so Fnet = Fmotor − mg = ma gives the force exerted by the motor as
Fmotor = m ( a + g ) = ( 650 kg ) ⎡⎣( 0.583 + 9.80) m s 2 ⎤⎦ = 6.75 × 103 N .
The average velocity of the elevator during this interval is
v f + vi
v=
= 0.875 m s , so the average power input from the motor during
2
this time is
⎛ 1 hp ⎞
℘= Fmotor v = 6.75 × 103 N ( 0.875 m s ) ⎜
= 7.92 hp .
⎝ 746 W ⎟⎠
(
)
(b) When the elevator moves upward with a constant speed of v = 1.75 m s , the
upward force exerted by the motor is Fmotor = mg and the instantaneous
power input from the motor is
⎛ 1 hp ⎞
℘= ( mg ) v = ( 650 kg ) 9.80 m s 2 (1.75 m s ) ⎜
= 14.9 hp .
⎝ 746 W ⎟⎠
(
)
A 24.2-kg child on a 1.95-m-long swing is released from rest when the ropes of the swing make an angle of
with the vertical.
(a) Neglecting friction, find the child's speed at the lowest position.
(b) If the actual speed of the child at the lowest position is 1.50 m/s, what is the mechanical energy lost due
to friction?
Choose PEg = 0 at the level of the bottom of the arc. The child’s
5.42 (a)
initial vertical displacement from this level is
y i = ( 2.00 m )(1 − cos 30.0°) = 0.268 m .
In the absence of friction, we use conservation of mechanical energy as
( KE + PE ) = ( KE + PE ) , or 12 mv
g
g
f
(
i
2
f
+ 0 = 0 + mgy i , which gives
)
v f = 2 gy i = 2 9.80 m s 2 ( 0.268 m ) = 2.29 m s .(b)
With a non-
conservative force present, we use
(
Wnc = KE + PEg
) − ( KE + PE ) = ⎛⎜⎝ 12 mv
f
g
i
2
f
⎞
+ 0⎟ − ( 0 + mgy i ) , or
⎠
⎛ v 2f
⎞
Wnc = m ⎜ − gy i ⎟
⎝ 2
⎠
⎡ ( 2.00 m s ) 2
⎤
= ( 25.0 kg ) ⎢
− 9.80 m s 2 ( 0.268 m ) ⎥ = −15.6 J
2
⎢⎣
⎥⎦
(
)
Thus, 15.6 J of energy is spent overcoming friction