A 377-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is 0.600 m, and its moment of inertia about its rotation axis is 0.800MR^2. Friction does work on the wheel as it rolls up the hill to a stop, a height h above the bottom of the hill; this work has absolute value 2700 J .
(a) Calculate h.

Explanation:

We shall apply law of conservation of mechanical energy to solve the problem

Initially wheel has rotational kinetic energy + linear kinetic energy

= 1 / 2 I ω² + 1/2 m v² , I is moment of inertia , m is mass , ω is angular velocity and v is linear velocity

=  1 / 2 I ω² + 1/2 x 2 Ix v² / r² (  I = 1/2 m r² )

= 1 / 2 I ω² +  Ix ω² ( v / r = ω )

= 3/2 I ω²

Given I = .8MR²

= .8 x 377/9.8 x .6²

= 11.07 ,

ω = 25 rad /s

Total kinetic energy = 3/2 x 11.07 x 25²

= 10378  J

energy lost due to work done by friction

= 10378 - 2700 = 7678 J

This energy will help it to scale height

mgh = 7678 , h is ultimate height achieved

h = 7678 / mg

= 7678 / 377

= 20.36 m .

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.

i think the answer is a, hope this is right, that would be my best guess.