A crate of fruit with a mass of 39.5 kg and a specific heat capacity of 3650 J/(

Question

A crate of fruit with a mass of 39.5 kg and a specific heat capacity of 3650 J/(kg*K) slides 7.20 m down a ramp inclined at an angle of 39.2 degrees below the horizontal.

If the crate was at rest at the top of the incline and has a speed of 2.00 m/s at the bottom, how much work W-final 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?


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 DeltaT?

thank you thank you!

Explanation / Answer

a)W=-E

W=PE1+KE1-(KE2+PE2)

W=PE1-KE2

W=mgh-(1/2)mv2

W=(39.5)(9.81)(7.2sin39.2)-(1/2)(39.5)(2)2

W=1682.5 joules

b)h=W

T=W/(cm)

T=(1682.5)/(39.5*3650)

T=.01167 K

Hope that helps

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