A crate of fruit with mass 35.0 kg and specific heat capacity 3650 J/Kg . K slides down a ramp inclined 36.9 degrees below the horizontal. the ramp is 8.00 m long. a) If the crate was at the rest the top of the incline and has a speed of 2.5o m/s the bottom how much work was done on the crate by friction? b) If an amount of heat equal to the work done by friction goes into the crate of fruit and the fruit reacches a uniform final temperature, What is the temperature change?

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A crate of fruit with mass 35.0 kg and specific heat capacity 3650 J/Kg . K slides down a ramp inclined 36.9 degrees below the horizontal. the ramp is 8.00 m long.  a) If the crate was at the rest the top of the incline and has a speed of 2.5o m/s  the bottom how much work was done on the crate by friction?  b) If an amount of heat equal to the work done by friction goes into the crate of fruit and the fruit reacches a uniform final temperature, What is the temperature change?

Jozlyn · Accepted Answer

a) Loss in P.E = work done against friction + gain in K.E  mgh = work done against friction -12mv2  work done against friction =35.0⋅9.8⋅8sin⁡36.9−0.5⋅35.0⋅2.502  = 1540 J  b) mc(temperature change) = 1540  35.0⋅3650⋅T=1540  T=0.0120 deg

A crate of fruit with mass 35.0 kg and specific heat capacity 3650 J/Kg . K slides down a ramp inclined 36.9 degrees below the horizontal. the ramp is

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