Question

A solid sphere of weight 36.0 N rolls up an incline at an angle of 30.0 . At the bottom of the incline the center of mass of the sphere has a translational speed of 4.90 m/s.(a) What is the kinetic energy of the sphere at the bottom of the incline? (b) How far does the sphere travel up along the incline? (c) Does the answer to (b)depend on the sphere’s mass?

Answer 1

36 N = (mass)(gravity)

1) The initial kinetic energy is both translational and rotational.

Initial KE = (1/2)MV^2 + (1/2)Iw^2

The moment of inertia(I) for a solid sphere is 2/5mR^2.

w = V/R

KE = 1/2mV^2 + 1/2(2/5mR^2)(V/R)^2

KE = 1/2mV^2 + 1/5mV^2

KE = 1/2(36 N/9.8 m/s^2)(4.9 m/s)^2 + 1/5(36 N/9.8 m/s^2)(4.9 m/s)^2 = 61.74 J

2) The initial kinetic energy at the bottom of the incline is equal to the potential energy at the top:

KE = mgh

61.74 J = (36N)h

h = 1.715 m

sin(theta) = h/d

sin(30) = (1.715 m)/d

d = 3.43 m up the incline

(c) Yes