Download Homework 05 - University of Utah Physics

Physics 1010, Spring 2007
Homework #5
SOLUTIONS
1. (Ch06, Q01) Block B has the greater amount of work done on it. Work = force × distance,
and block B moves twice as far. The mass of the blocks is irrelevant.
2. (Ch06, Q06) The tension in the string does no work on the ball. The force applied by
the string is perpendicular to the motion of the ball. Only forces parallel to the motion can
do work.
3. (Ch06, Q11) The kinetic energy of the friend is equal to the work done by the force exerted
by the boy. The fact that the boy receives an equal and opposite force from his friend —
and hence has word done on him — does not change the work done on the friend.
4. (Ch06, Q14) No! Work done is proportional to the change in kinetic energy, which is
proportional to the velocity squared:
1
W = F × d = mv 2
2
If the velocity of one ball is 2× as great, the work done on that ball is 22 = 4× as great.
5. (Ch06, Q22)
(a) Prior to its release, gravitational potential energy is added to the pendulum. It
gains potential energy P E = mgh as a result of being lifted through a height h.
(b) The pendulum’s kinetic energy is greatest when the pendulum is at the bottom of
its swing. The pendulum has no “height” at the bottom of the swing, so all its energy
must have been converted to kinetic.
(c) Potential energy is greatest when the pendulum is at either of the two end points of
its swing. At these positions, the pendulum momentarily comes to rest and hence has
no kinetic energy. All of its energy at the endpoints must be in the form of gravitational
potential energy.
6. (Ch06, E02) The definition of work is
W =F ×d
We can perform algebra on this expression to obtain
F =
160
W
=
= 40 Newtons
d
4
7. (Ch06, E04)
(a) For the force applied by the rope,
W = F × d = 180 × 2 = 360 Joules
(b) For the force applied by friction, since the force is directed opposite the direction of
motion,
W = F × d = −60 × 2 = −120 Joules
(c) The total work done is the sum of work done in (a) and (b), or 240 Joules.
8. (Ch06, E06)
(a) The work done is
W = F × d = 60 × 10 = 600 Joules
(b) The increase in kinetic energy is equal to the work done, or 600 Joules. (Note that
the mass of the object doesn’t matter!)
9. (Ch06, E10)
(a) The increase in potential energy is just the work done, 40 Joules.
(b) The elastic potential energy stored in a spring is given by
1
P E = kx2
2
where x is the displacement of the spring from its equilibrium (unstretched) position
and k is the spring constant. We can solve the equation above for k to obtain:
k=
2 × 40
2 × PE
=
= 2, 000 Newtons/meter
2
x
0.202
10. (Ch06, E12)
(a) Kinetic energy is given by
1
1
KE = mv 2 = × 0.2 × 42 = 1.6 Joules
2
2
(b) Ignoring air resistance, total mechanical energy will be conserved. This means that the
pendulum will rise until its potential energy equals the kinetic energy at the low point
in the swing.
P E = mgh = 1.6 Joules
solving for the height h, we obtain
h=
1.6
PE
=
= 0.816 meters
mg
0.2 × 9.8
(= 0.8 meters if g = 9.8 m/s2 )
11. (Ch06, E14)
(a) At the top of the hill, the potential energy of the sled and rider is
P E = mgh = 50 × 9.8 × 15 = 7, 350 Joules
(b) After the push, the total mechanical energy is the sum of the potential and kinetic
energies:
ME = P E + KE = 7, 350 + 1, 600 = 8, 950 Joules
(c) At the bottom of the hill, potential energy is equal to zero so all of the mechanical
energy becomes kinetic. Final KE = 8,950 Joules.