Download CHAPTER 8 SOLUTION FOR PROBLEM 9 (a) The only force that

CHAPTER 8
SOLUTION FOR PROBLEM 9
(a) The only force that does work as the flake falls is the force of gravity and it is a conservative
force. If Ki is the kinetic energy of the flake at the edge of the bowl, Kf is its kinetic energy at
the bottom, Ui is the gravitational potential energy of the flake-Earth system with the flake at the
top, and Uf is the gravitational potential energy with it at the bottom, then Kf + Uf = Ki + Ui .
Take the potential energy to be zero at the bottom of the bowl. Then the potential energy at the
top is Ui = mgr, where r is the radius of the bowl and m is the mass of the flake. Ki = 0 since
the flake starts from rest. Since the problem asks for the speed at the bottom, write 12 mv 2 for
Kf . The energy conservation equation becomes mgr = 12 mv 2 , so
0
2
v = 2gr = 2(9.8 m/s )(0.220 m) = 2.08 m/s .
√
(b) Note that the expression for the speed (v = 2gr) does not contain the mass of the flake.
The speed would be the same, 2.08 m/s, regardless of the mass of the flake.
(c) The final kinetic energy is given by Kf = Ki + Ui − Uf . Since Ki is greater than before, Kf
is greater. This means the final speed of the flake is greater.
CHAPTER 8
SOLUTION FOR PROBLEM 25
Information given in the second sentence allows us to compute the spring constant. Solve F = kx
for k:
270 N
F
=
= 1.35 × 104 N/m .
k=
x 0.02 m
(a) Now consider the block sliding down the incline. If it starts from rest at a height h above
the point where it momentarily comes to rest, its initial kinetic energy is zero and the initial
gravitational potential energy of the block-Earth system is mgh, where m is the mass of the
block. We have taken the zero of gravitational potential energy to be at the point where the block
comes to rest. We also take the initial potential energy stored in the spring to be zero. Suppose
the block compresses the spring a distance x before coming momentarily to rest. Then the final
kinetic energy is zero, the final gravitational potential energy is zero, and final spring potential
energy is 12 kx2 . The incline is frictionless and the normal force it exerts on the block does no
work, so mechanical energy is conserved. This means mgh = 12 kx2 , so
h=
(1.35 × 104 N/m)(0.055 m)2
kx2
=
= 0.174 m .
2
2mg
2(12 kg)(9.8 m/s )
If the block traveled down a length of incline equal to f, then f sin 30◦ = h, so f = h/ sin 30◦ =
(0.174 m)/ sin 30◦ = 0.35 m.
(b) Just before it touches the spring it is 0.055 m away from the place where it comes to rest
and so is a vertical distance h = (0.055 m) sin 30◦ = 0.0275 m above its final position. The
2
gravitational potential energy is then mgh = (12 kg)(9.8 m/s )(0.0275 m) = 3.23 J. On the other
2
hand, its initial potential energy is mgh = (12 kg)(9.8 m/s )(0.174 m) = 20.5 J. The difference is
its final kinetic energy: Kf = 20.5 J − 3.23 J = 17.2 J. Its final speed is
5
2Kf
2(17.2 J)
=
= 1.7 m/s .
v=
m
12 kg
CHAPTER 8
SOLUTION FOR PROBLEM 39
(a) The magnitude of the force of friction is f = µk N , where µk is the coefficient of kinetic
friction and N is the normal force of the surface on the block. The only vertical forces acting on
the block are the normal force, upward, and the force of gravity, downward. Since the vertical
component of the block’s acceleration is zero, Newton’s second law tells us that N = mg, where
m is the mass of the block. Thus f = µk mg. The increase in thermal energy is given by
∆Eth = f f = µk mgf, where f is the distance the block moves before coming to rest. Its value is
2
∆Eth = (0.25)(3.5 kg)(9.8 m/s )(7.8 m) = 67 J.
(b) The block had its maximum kinetic energy just as it left the spring and entered the part of the
surface where friction acts. The maximum kinetic energy equals the increase in thermal energy,
67 J.
(c) The energy that appears as kinetic energy is originally stored as the potential energy of the
compressed spring. Thus ∆E = 12 kx2 , where k is the spring constant and x is the compression.
Solve for x:
5
2∆E
2(67 J)
=
= 0.46 m .
x=
k
640 N/m
CHAPTER 8
HINT FOR PROBLEM 11
Use conservation of mechanical energy. When the car is a distance y above ground level and is
traveling with speed v the mechanical energy is given by 12 mv 2 + mgy, where m is the mass of
the car. Write this expression for the initial values and for the values when the car is at another
point (A, B, C, or the point where it stops on the last hill). Equate the two expressions to each
other and solve for the unknown (either the speed or the height above the ground). Notice that
the mass of the car cancels from the conservation of energy equation.
J
ans: (a) 17.0 m/s; (b) 26.5 m/s; (c) 33.4 m/s; (d) 56.7 m; (e) all the same
o
CHAPTER 8
HINT FOR PROBLEM 29
Take the gravitational potential energy to be zero when the whole cord is stuck to the ceiling.
Now compute the potential energy when the cord is hanging by one end. Consider an infinitesimal
segment of cord with mass dm a distance y from the ceiling. The potential energy associated
with this segment is dU = −gy dm. If the length of the segment is dy, then dm = (dy/L)M ,
where L is the length of the cord and M is its mass. The total potential energy is the sum over
segments:
8 L
M
gy dy .
U =−
L
0
Evaluate the integral.
J
ans: −18 mJ
o
CHAPTER 8
HINT FOR PROBLEM 43
(a) The work done by the spring force is given by Ws = − 12 kd2 , where d is the spring compression
and k is the spring constant.
(b) The change in the thermal energy is given by ∆Eth = f d, where f is the magnitude of the
frictional force. This is the product of the magnitude of the normal force of the floor on the
block and the coefficient of kinetic friction. Use Newton’s second law to obtain the magnitude
of the normal force.
(c) Use the energy equation W = ∆Emec + ∆Eth , where W is the work done by external forces
and ∆Emec is the change in the mechanical energy. Take the system to be composed of the block
and the spring. Then no external forces do work and the mechanical energy is the sum of the
kinetic energy of the block and the potential energy stored in the spring. Solve for the initial
kinetic energy and then the initial speed.
J
ans: (a) −0.90 J; (b) 0.46 J; (c) 1.0 m/s
o