A crate of fruit with a mass of 33.0kg and a specific heat capacity of 3500J/(kg

Question

A crate of fruit with a mass of 33.0kg and a specific heat capacity of 3500J/(kg?K) slides 9.00m down a ramp inclined at an angle of 38.9degrees 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/s2 for the acceleration due to gravity and express your answer in joules.

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 ?T?

Explanation / Answer

Here , Using Conservation of energy ,

KE = PE + Wf

0.5mv^2 = Wf + mgh

0.5 * 33 * 2.8^2 = Wf + 33*9.8*9*sin(38.9)

solving for Wf

Wf = -1698.4 J

Wf work done by friction is -1698.4 J

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