Question

A solid sphere of weight 42.0 N rolls up an incline at an angle of 36.0°. At the bottom of the incline the center of mass of the sphere has a translational speed of 5.10 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

Answer:

Part a)

$KE = 77.95 J$

Part b)

$L = 3.16 m$

Part c)

distance L is independent of the mass of the sphere

Explanation:

Part a)

As we know that rotational kinetic energy of the sphere is given as

$KE = \frac{1}{2}I\omega_2 + \frac{1}{2}mv^2$

so we will have

$KE = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2$

so we will have

$KE = \frac{1}{5} mv^2 + \frac{1}{2}mv^2$

$KE = \frac{7}{10} mv^2$

$KE = \frac{7}{10}(\frac{42}{9.81})(5.10^2)$

$KE = 77.95 J$

Part b)

By mechanical energy conservation law we know that

Work done against gravity = initial kinetic energy of the sphere

So we will have

$mgLsin\theta = KE$

$\frac{42}{9.81}(9.81)L sin36 = 77.95$

$L = 3.16 m$

Part c)

by equation of energy conservation we know that

$\frac{7}{10}mv^2 = mgL sin\theta$

so here we can see that distance L is independent of the mass of the sphere