Question

# A crate of fruit with a mass of 35.0 kg and a specific heat capacity of...

A crate of fruit with a mass of 35.0 kg and a specific heat capacity of 3800 J/(kg⋅K) slides 8.40 m down a ramp inclined at an angle of 36.1 degrees below the horizontal.

Part A: If the crate was at rest at the top of the incline and has a speed of 2.65 m/s at the bottom, how much work Wf was done on the crate by friction? Use 9.81 m/s2 for the acceleration due to gravity and express your answer in joules.

Given data:
mass = 35 kg
heat capacity = 3800 J/ Kg.K
sliding down = 8.40 m
angle = 36.1 degrees
speed = 2.65 m/s
g= 9.81 m/s^2

Work done by friction can be found as the difference of potential energy and kinetic energy.

Potential energy change: mgh = mgs sin 36.1°.
Kinetic energy change: ½ mv².
Work done by friction (on crate): mgs sin 36.1° − ½ mv²
W = 35*(9.8 *8.40* sin 36.1 − ½ * 2.65²) = 1574 J.

For a frictionless incline surface, potential energy change = kinetic energy change, along with null friction work. In the present case, however,
kinetic energy is only about 7% of potential energy: almost 93% of
energy is converted to heat.