A crate of fruit with a mass of 38.5kg and a specific heat capacity of 3650J/kg*

Question

A crate of fruit with a mass of 38.5kg and a specific heat capacity of 3650J/kg*K) slides 7.80m down a ramp inclined at an angle of 39.8degrees below the horizontal. If the crate was at rest at the top of the incline and has a speed of 2.80m/s at the bottom, how much work Wf was done on the crate by friction? Use 9.81m/s^2 for the acceleration due to gravity and express your answer in joules.

Explanation / Answer

The vertical height of the crate (h) = 7.80 x sin37* = 7.80 x 0.60 = 4.68m PE of the crate = mgh =>PE = 33 x 9.8 x 4.68 = 1513.50 J The KE of the crate at the bottom = 1/2mv^2 =>KE = 1/2 x 33 x (3)^2 = 148.50 J Thus by the law of energy conservation, =>W(f) = PE - KE =>W(f) = 1513.50 - 148.50 = 1365 J

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