A crate of fruit with a mass of 40.0 kg and a specific heat capacity of 3800 J/(

Question

A crate of fruit with a mass of 40.0 kg and a specific heat capacity of 3800 J/(kgK) slides 7.50 m down a ramp inclined at an angle of 36.0 degrees below the horizontal.

Part A

If the crate was at rest at the top of the incline and has a speed of 2.10 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

A)The height of the ramp is

h = 7.5 x sin36 = 4.41 m

Intial energy = Ei = m g h = 40 x 9.8 x 4.41 = 1728.72 J

Final energy = 1/2 m v^2 = 0.5 x 40 x 2.1^2 = 88.2 J

Wf = Ef - Ei = 88.2 - 1728.72 = -1640.52

Hence, Wf = -1640.52 J

B)We know that,

Q = m c delta T

delta T = Q/mc = 1640.52/40 x 3800 = -0.011 deg C

Hence, delta T = 0.011 deg C

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