A crate of fruit with a mass of 36.5 kg and a specific heat capacity of 3700 J/(

Question

A crate of fruit with a mass of 36.5 kg and a specific heat capacity of 3700 J/(kgK) slides 7.70 m down a ramp inclined at an angle of 37.7 degrees below the horizontal.

If the crate was at rest at the top of the incline and has a speed of 3.00 m/s at the bottom, how much work Wf was done on the crate by friction?

Use 9.81 m/s2 for the acceleration due to gravity and express your answer in joules.

Part 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 T?

Explanation / Answer

crate of fruit has a KE at bottom of ramp,

K.E = 1/2(36.5)(3)^2 = 164.25 J

crate of fruit at top of ramp (not sliding) has a total energy,

GPE = mgh = 36.5(9.81)(7.70)(sin 37.7°) = 1686 J

work done by friction on crate of fruit = 1686 - 164.25 = 1521.75 J Ans(a)

Q = 1521.75 = m(cp)T = 36.5(3700)T

T = 1521.75/135050

T = 0.01126 °C .......Ans(b)

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