CHAPTER 5
WORK, ENERGY, AND POWER
1)
A 900N crate rests on the floor. How much work is
required to move it at a constant speed (a) 6.0m along the floor against a
frictional force of 180 N, and (b) 6.0m vertically?
SOLUTION:
(a) Because there is no acceleration, the
horizontal applied force must have the same magnitude as
W
= FDx = (180 N)(6.0 m) =1.1 ´ 10^{3} J.
(b) Because there is no acceleration, the
vertical applied force must have the same magnitude as

2)
At room temperature, an oxygen molecule, with a mass of
5.31 x 10^{26} kg, typically has a kinetic energy of about 6.21 x 10^{21}
J. How fast is it moving?
SOLUTION:
We
find the speed from
ke = ½ mv^{2};
6.21 ´ 10^{–21} J = ½ (5.31 ´ 10^{–26} kg)v^{2},
which gives v =484 m/s.
SOLUTION:
The
work done on the electron decreases its kinetic energy:
W
= Dke
= ½mv^{2} – ½mv_{0}^{2} = 0 – ½ (9.11 ´ 10^{–31} kg)(1.90 ´ 10^{6} m/s)^{2} = – 1.64 ´ 10^{–18} J.
SOLUTION:
The
potential energy of the spring is zero when the spring is not stretched (x=
0). For the stored potential energy, we have
pe = ½kx_{f}^{2}–
0;
25
J = ½ (440 N/m) x_{f}^{2} – 0, which gives x_{f}=0.34 m.
SOLUTION:
For
the potential energy change we have
Dpe
= mg Dy = (64 kg)(9.80
m/s^{2})(4.0 m) =2.5 ´ 10^{3}
J.
SOLUTION:
We
choose the potential energy to be zero at the bottom (y = 0). Because
there is no friction and the normal force does no work, energy is conserved, so
we have
e = ke_{i }+ pe_{i }= ke_{f }+ pe_{f };
½mv_{i}^{2}
+ mgy_{i} = ½mv_{f}^{2} + mgy_{f };
½m(0)^{2}
+ m(9.80 m/s^{2})(125 m) = ½mv_{f}^{2} + m(9.80
m/s^{2})(0), which gives v_{f} =49.5
m/s.
This is 180 km/h!
It is a good thing there is friction on the ski slopes.
SOLUTION:
The
amount of work required is the increase in potential energy: W = mg Dy.
We find the time
from
P
= W/t = mg Dy/t;
1750
W = (285 kg)(9.80 m/s^{2})(16.0 m)/t, which gives t =25.5 s.
SOLUTION:
We
find the equivalent force exerted by the engine from
P
= Fv;
(18
hp)(746 W/hp) = F(90 km/h)/(3.6 ks/h), which gives F = 5.4 ´ 10^{2} N.
At constant speed,
this force is balanced by the average retarding force, which must be5.4 ´ 10^{2} N.
1) What
is the minimum work needed to push a 1000kg car 300 m up a 17.7^{0}
incline? (a) ignore friction. (b) Assume the effective coefficient of friction
is 0.25.
SOLUTION:
The
minimum work is needed when there is no acceleration.
(a) From the force diagram, we write SF = ma:
ycomponent:
F_{N} – mg cos q
= 0;
xcomponent:
F_{min} – mg sin q
= 0.
For
a distance d along the incline, we have
W_{min }= F_{min}d
cos 0° = mgd sin q
(1)
=
(1000 kg)(9.80 m/s^{2})(300 m) sin 17.5°
=8.8 ´ 10^{5} J.
(b) When there is friction, we have

xcomponent:
F_{min} – mg sin q
– m_{k}F_{N}
= 0,or
F_{min}
= mg sin q + m_{k}mg cos q= 0,
For
a distance d along the incline, we have
W_{min }= F_{min}d
cos 0° = mgd (sin q
+ m_{k} cos q)(1)
=
(1000 kg)(9.80 m/s^{2})(300 m)(sin 17.5° + 0.25 cos 17.5°) =1.6 ´ 10^{6} J.
SOLUTION:
On
a level road, the normal force is mg, so the kinetic friction force
is m_{k}mg .
Because it is the (negative) work of the friction force that stops the car, we
have
W
= Dke;
– m_{k}mg d = ½mv^{2}
– ½mv_{0}^{2};
–
(0.42)m(9.80 m/s^{2})(88 m) = – ½mv_{0}^{2},
which gives v_{0} =27 m/s (97 km/h or
60 mi/h).
Because
every term contains the mass, it cancels.
SOLUTION:
(a) With the reference level at the ground,
for the potential energy change we have
Dpe
= mg Dy = (55 kg)(9.80
m/s^{2})(3100 m – 1600 m) = 8.1 ´ 10^{5}
J.
(b) The minimum work would be equal to the
change in potential energy:
W_{min}
=Dpe
= 8.1 ´10^{5} J.
(c)
Yes, the actual work will
be more than this. There will be additional work required for any
SOLUTION:
We
choose the potential energy to be zero at the ground (y = 0).Energy is
conserved, so we have
e = ke_{i} + pe_{i }= ke_{f} + pe_{f };
½mv_{i}^{2}
+ mgy_{i} = ½mv_{f}^{2} + mgy_{f };
½m(185
m/s)^{2} + m(9.80 m/s^{2})(265 m) = ½mv_{f}^{2}
+ m(9.80 m/s^{2})(0), which gives v_{f} = 199 m/s.
Note that we have
not found the direction of the velocity.
SOLUTION:

With y = 0
at the bottom of the circle, we call the start
At the top of the
circle we have the forces mg and F_{N}, both
mg
+ F_{N} = mv_{C}^{2}/r.
The minimum value
of F_{N} is zero, so the minimum speed at C
v_{C}_{min}^{2}
= gr.
From energy
conservation for the motion from A to C we have
ke_{A} + pe_{A} = ke_{C} + pe_{C };
0
+ mgh = ½mv_{C}^{2} + mg(2r),
thus the minimum
height is found from
gh = ½v_{C}_{min}^{2} + 2gr = ½gr + 2gr, which gives h = 2.5r.
SOLUTION:
On
the level the normal force is F_{N}= mg, so the friction
force is F_{fr} = µ_{k}mg.
For the
workenergy principle, we have
W_{NC }=
Dke
+ Dpe
= (½mv_{f}^{2} – ½mv_{i}^{2}) + mg(h_{f}
– h_{i});
F(L_{1}
+ L_{2}) –µ_{k}mg L_{2} =
(½mv_{f}^{2} – 0) + mg(0 – 0);
(350
N)(15 m + 15 m) – (0.30)(90 kg)(9.80 m/s^{2})(15 m) = ½ (90 kg)v_{f}^{2},
which gives v_{f}
=12 m/s.
SOLUTION:
The
work done increases the potential energy of the player. We find the power from
P = W/t = Dpe/t
= mg(h_{f} – h_{i})/t
= (105 kg)(9.80 m/s^{2})[(140 m) sin 30° – 0]/(61 s) = 1.2 ´ 10^{3} W(about 1.6 hp).
1) A
220kg load is lifted 21.0m vertically with an acceleration a = 0.150 g (where
g = 9.80 m/s^{2}) by a single cable. Determine (a) the tension in the
cable, (b) the net work done on the load, (c) the work done by the cable on the
load, (d) the work done by gravity on the load, (e) the final speed of the load
assuming that it started from rest.
SOLUTION:
(a) From the force diagram we write SF_{y} = ma_{y}:

F_{T }–
mg = ma;
F_{T }–
(220 kg)(9.80 m/s^{2}) = (220 kg)(0.150 m/s^{2}),
which
gives F_{T} =2.48 ´ 10^{3}
N.
(b) The net work is done by the net force:
W_{net }= F_{net}h
= (F_{T }– mg)h
=
[2.48 ´10^{3} N – (220 kg)(9.80
m/s^{2})](21.0 m) = 6.79 ´10^{3}
J.
(c) The work is done by the cable is
W_{cable }= F_{T}h
=
(2.48 ´10^{3} N)(21.0 m) =5.21 ´10^{4} J.
(d) The work is done by gravity is
W_{grav }= – mgh
=
– (220 kg)(9.80 m/s^{2})(21.0 m) =– 4.53 ´10^{4}
J.
Note
that W_{net} = W_{cable} + W_{grav}.
(e) The net work done on the load increases
its kinetic energy:
W_{net}
= Dke
= ½mv^{2} – ½mv_{0}^{2} ;
6.79 ´10^{3} J = !(220 kg)v^{2} –0, which gives
v =7.86 m/s.
SOLUTION:
The
maximum acceleration will occur at the maximum compression of the spring:
kx_{max}
= Ma_{max}= M(5.0g), which gives x_{max}
= 5.0Mg/k.
For the motion
from reaching the spring to the maximum compression of the spring,
we use energy
conservation:
ke_{i} + pe_{springi }= ke_{f} + pe_{springf};
½Mv^{2}
+ 0 = 0 + ½kx_{max}^{2}.
When we use the
previous result, we get
½Mv^{2}
= ½k(5.0Mg/k)^{2}, which gives
k
= 25Mg^{2}/v^{2} = 25(1200 kg)(9.80 m/s^{2})^{2}/[(100
km/h)/(3.6 ks/h)]^{2} =3.7 ´10^{3}
N/m.
SOLUTION:
On
the level the normal force is F_{N}= mg, so the friction
force is F_{fr} = µ_{k}mg.
The block is at
rest at the release point and where it momentarily stops before turning back.
For the
workenergy principle, we have
W_{NC }=
Dke
+ Dpe
= (½mv_{f}^{2} – ½mv_{i}^{2}) +
(½kx_{f}^{2} – ½kx_{i}^{2});
–µ_{k}mg
L = (0 – 0) + ½k(x_{f}^{2} – x_{i}^{2}
);
–
µ_{k}(0.520 kg)(9.80 m/s^{2})(0.050 m + 0.023 m) = ½
(180 N/m)[(0.023 m)^{2} – (– 0.050 m)^{2}],
which gives µ_{k}
=0.48.
SOLUTION:

From
the force diagram for the car, we have:
xcomponent: F – F_{fr}= mg
sin q.
Because the power
output is P = Fv, we have
(P/v)
– F_{fr}= mg sin q.
The maximum power
determines the maximum angle:
(P_{max}/v)
– F_{fr}= mg sin q_{max };
(120
hp)(746 W/hp)/[(70 km/h)/(3/6 ks/h)] – 600 N =
(1000
kg)(9.80 m/s^{2}) sin q_{max },
which gives
sin q_{max} =
0.409,orq_{max} = 24°.
SOLUTION:
The
energy of the mass is purely potential before it was dropped.
E_{i} =
mgy_{i} = (0.10kg)(9.80 m/s^{2})(1.0 m) = 0.98 J
The
energy is purely potential once the spring is fully compressed, however it has
both gravitational potential and elastic potential energy.
E_{f} =
mgy_{f} + ½ ky^{2}_{f} = (0.10 kg)(9.80 m/s^{2})(0.25
m) + (0.5)k(0.25 m)^{2} = 0.245 J + .03125 k
Conservation
of energy applies:
0.98 J = 0.245 J
+ .03125 k
k = 39.2 N/m