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A crate of fruit with a mass of 37.5 kg and a specific heat capacity of 3800 J/(kg⋅K) slides 7.60 m down a ramp inclined at an angle of 38.5 degrees below the horizontal.

If the crate was at rest at the top of the incline and has a speed of 2.35 m/s at the bottom, how much work Wf was done on the crate by friction? Use 9.81 m/s^2 for the acceleration due to gravity and express your answer in joules.

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2 answers
  1. initial PE-finalKE=workdonebyfriction
    =mg*7.60*sin38.5- 1/2 m 2.35^2

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    bobpursley
  2. 1904 J

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