A crate of fruit with a mass of 45.0 kg and a specific heat capacity of 3600 J/(
ID: 1522004 • Letter: A
Question
A crate of fruit with a mass of 45.0 kg and a specific heat capacity of 3600 J/(kg×K) slides 10.0 m down a ramp inclined at an angle of 37.0 degrees below the horizontal. The kinetic coefficient of friction between the crate and ramp is 0.250. (a) What is the speed of the crate at the bottom? (b) If an amount of heat equal to the magnitude of the work done by friction is absorbed by the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change DT?
No clue on how to do this one. Answers are A) 8.88 m/s B) .00544 degrees C
Any help is appreciated!
Explanation / Answer
a) Final kinetic energy = Loss in potential energy - work done by friction
0.5 mv^2 = mgh sin theta - u mg cos theta
0.5 * 45 * v^2 = 45 * 9.8 * 10 * sin37 - 0.250 * 45 * 9.8 * 10 * cos37
solving for v = 8.878 m/sec
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b) amount of heat equal to the magnitude of the work done by friction
u mg cos theta = mQ DT
0.250 * 45 * 9.8 * 10 * cos37 = 45 * 3600 * T
solving for T
T = 0.00543 degree celsius
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