PHYSICS QUESTION

Question

A crate of fruit with a mass of 34.5  and a specific heat capacity of 3800  slides 8.20  down a ramp inclined at an angle of 37.0  below the horizontal.

If the crate was at rest at the top of the incline and has a speed of 2.60 

 at the bottom, how much work  was done on the crate by friction?

Answer 1

The work done by friction should equal the difference between the actual kinetic energy at the bottom, and the expected kinetic energy based on the initial height.

Actual kinetic energy:

K = mv²/2

Expected kinetic energy:

K₀ = mgh = mgLsinθ

Work done by friction (should come out negative):

W = K - K₀ = mv²/2 - mgLsinθ = m(v²/2 - gLsinθ)

Plug in the values of m, v, g, L, and θ, in correct units. By the way, you did not provide the units for most of these quantities.