Answers for Problem Set 8

Chapter 8

   (10 a) W=mg(h-R)

   (10 b) W=mg(h-2R)

   (10 c) U=mgh

   (10 d) U=mgR

   (10 e) U=mg(2R)

   (39 a) F = - 8 m g 1_x - m g 1_y

   (39 b) h = ( 5/2 ) R

40. 

The maximum tension is T = m g + 2 m g d / L = 933 Newtons, which is OK.

44. 

(a) d = [ ( 1/2) K x_f² - m g | x_f | sin (theta) ] / [ m g sin (theta) ] = 0.40 m

(b) x₂ = m g sin (theta) / K = -0.04 m

46.

This answer is entirely graphical and will be on the solutions page only.

49.

(a) There is a turning point at 0.08 nm.

(b) There are turning points at 0.15 and 0.3 nm and the particle can only be between these two points.

(c) Potential energy = -1 x 10 ^-19 joules

(d) Kinetic energy = 2 x 10 ^-19 joules

(e) Force 10 ^-9 newtons, to the left on m and to the right on M.

(f) Repulsive from near r = 0 to about r = 0.2 nm.

(g)
Attractive outside r = 0.2 nm.

(h)
Zero at r = 0.2 nm.

75. 

(a)
v₁ = sqrt { 2 g d [ sin (theta) - mu_k cos (theta) ] }
v₁ = 18 ft / sec

(b)
x = v₁ ² / [ 2 mu_k g ]
x = 18 ft.