A crate of fruit with a mass of 30.5 kg and a specific heat capacity of 3550 J/(

Question

A crate of fruit with a mass of 30.5 kg and a specific heat capacity of 3550 J/(kg?K) slides 8.20 m down a ramp inclined at an angle of 39.1 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.

A crate of fruit with a mass of 30.5 kg and a specific heat capacity of 3550 J/(kg K) slides 8.20 m down a ramp inclined at an angle of 39.1 degrees below the horizontal.

Explanation / Answer

decrease in height = L* sin 39.1
= 8.2* sin 39.1
= 5.17 m

decrease in potential energy = m*g*decrease in height
= 30.5*9.8*5.17
=1545.8 J

increase in kinetic energy = 0.5*m*vf^2
= 0.5*30.5*3^2
= 137.25 J

difference between them is the work done by friction and it should be negative

work done by friction = 137.25 - 1545.8 = - 1408.55 J
Answer: - 1408.55 J

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