Preparation problems

8-44. Figure 8-32 (p. 199) shows the potential energy curve for a certain particle. Find the force on the particle at each of the curve segments shown.

Solution:

We start by noting that since the potential energy is the integral of the force acting on a particle over a distance, the force must be given by the derivative of the potential energy curve, or

Fx = dU
dx
So to find the force for each segment, we must find the slope of that segment. Using rise over run for each segment, we get a force for each segment of:

(a) -2 N

(b) 0 N

(c) 8 N

(d) 1 N

(e) -4 N

(f) 0 N

8-51. A basketball dropped from a height of 2.00 m rebounds to a maximum height of 1.12 m. What fraction fo the ball's initial energy is lost to nonconservative forces? Take the zero of potential energy at the floor.

Solution:

We want to find the fraction of total mechanical energy lost. This is given by

h = Ei - Ef
Ei
 = 1 - Ef
Ei

Since at the time it is dropped, the ball has no kinetic energy, and at the highest point of its motion after it rebounds it likewise has no kinetic energy, the total mechanical energy at these points is given by the potential energy. Thus

h
=
1 - Ef
Ei
=
1 - Uf
Ui
=
1 - mgyf
mgyi
=
1 - yf
yi
=
1 - 1.12  m
2.0  m
h
=
0.44

8-64. The Sun's total power output is 3.85 ×10+26 W. What is the associated rate at which the Sun loses mass?

Solution:

First, I must point out that there is a typographical error in the book, and the exponent of the Sun's power output is +26, not -26. This said, we can now relate the energy output by the Sun each second to a quantity of mass lost using Einstein's famous relationship

E = m c2
Each second, the sun produces 3.85 ×10+26 J. The mass equivalence of this amount of energy is
E
=
m c2
m
=
E
c2
=
3.85 ×10+26  J
(3.0 ×108  m/s)2
m
=
4.27 ×109  kg
So the rate at which the Sun loses mass is 4.27 ×109  kg/s.

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