Question

A solid, uniform sphere with a mass of 2.0 kg is rolling from rest down an incline plane from the top of the plane. The incline plane makes an angle of 20◦ with the horizontal and has a height of 2.0 m. At the bottom of the incline plane, the surface levels out to a frictionless horizontal surface. A spring with a spring constant of k is located 5.0 m down the horizontal surface. If the spring is compressed by 5.0 cm upon impact with the sphere, what is the spring constant?

Answer 1

By energy conservation from top of incline to bottom at the plane,

Potential energy at the top = translation KE + rotational KE

mgh = 0.5mv^2 + 0.5iw^2 where m is mass, i is moment of inertia, w is angular velocity which is equal to v/r, v is velocity.

mgh = 0.5mv^2 + 0.5*2/5mr^2*w^2

gh = 0.5v^2 + 0.2 v^2

v = sqrt(gh/0.7) = sqrt(9.8*2/0.7) = 5.29 m/s

The translational KE gets converted to spring potential energy,

0.5kx^2 = 0.5mv^2

0.5*k*0.05^2 = 0.5*2*5.29^2

k = 22387 N/m answer