A ball is launched up an inclined wall by a spring The sprin

A ball is launched up an inclined wall by a spring. The spring constant of 3,400 N/m and is initially compressed 0.25 m. The ball has a mass of 1.3kg, a radius of 0.07m, and a mass moment of Inertia of 0.00817 kg-m^2. You can assume the ball rolls along the surface without sliding. The wall is inclined at an angle of 25degree. What is the velocity of the ball after it has traveled for  3.8 m along the inclined wall?

Solution

Here,

k = 3400 N/m

x 0.25 m

m = 1.3 Kg

r = 0.07 m

I = 8.17 *10^-3 Kg.m^2

theta = 25 degree

let the speed of ball is v

Using work energy theorum

0.5 * 1.3 * v^2 + 0.5 * I * (v/r)^2 = 0.5 * k * x^2 - m * g * L * sin(theta)

0.5 * 1.3 * v^2 + 0.5 * 8.17 *10^-3 * (v/0.07)^2 = 0.5 * 3400 * 0.25^2 - 1.3 * 9.8 * 3.8 * sin(25)

solving for v

v = 7.604 m/s

the velocity of ball after travelling 3.8 m up the incline is 7.604 m/s