A 1.65 kg wood block os released from rest on a frictionless ramp so that it beg

Question

A 1.65 kg wood block os released from rest on a frictionless ramp so that it begins to slide down the ramp. Thw ramp surface is sloped at angle 60.3 degrees from horizontal.

After it slides down a distance of 1.79m measured along tje ramp, how much total work has been done on the block (in Joules)?

Somebody measures the speed of the block after it has gone the 1.79m down the ramp. Predict what speed they should measure.

Another person places the wooden block back at the top of the ramp, but tjis time starts the blocl with a push downhill. With an initial speed of 2.00 m/s the block slides down the ramp 1.79m, what speed should they measure for the block?

Explanation / Answer

a. Work done total = W = mgh sin theta

W = 1.65 * 9.81* 1.79 * sin 60.3

W = 25.16 Joules

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KE = PE

0.5m v^2 = 25.16

v^2 = 2* 25.16/1.65

v = 5.522 m/s

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again change in KE = PE

0.5* m (V^2 - 2^) = 25.16

0.5 * 1.65 * (v^2 -4) = 25.16

v = 25.16/(0.5 * 1.65) + 4

v = 5.87 m/s

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