Question

Suppose that a ball decelerates from 8.0 m/s to a stop as it rolls up a hill, losing 10% of its kinetic energy to friction. Determine how far vertically up the hill the ball reaches when it stops. Show your work.

Answer 1

Hi there!

Assuming the ball is hollow and spherical:

`I = (2)/(3)MR^2`

Since the ball is both rolling and moving linearly, the total kinetic energy is comprised of both ROTATIONAL and TRANSLATIONAL kinetic energy.

`KE_T = KE_R + KE_(TR)`

Recall the equations for both:

`KE_(TR) = (1)/(2)mv^2\\\\KE_R = (1)/(2)I\omega^2`

We can rewrite the rotational kinetic energy using linear velocity using the following relation:

`\omega = (v)/(r)`

`KE_R = (1)/(2)((2)/(3)mR^2((v^2)/(R^2)))\\\\KE_R = (1)/(3)mv^2`

We can now represent the situation as a summation of energies:

`.9(KE_R + KE_(TR)) = PE\\\\.9((1)/(2)mv^2 + (1)/(3)mv^2) = mgh \\\\.9((5)/(6)mv^2) = mgh`

Cancel out the mass and rearrange to solve for height:

`(3)/(4)v^2 = gh \\\\h = (3v^2)/(4g)`

Plug in the given velocity and g = 9.8 m/s².

`h = (3(8^2))/(4(9.8)) \approx \boxed{4.9 m}`