A 4.0kg block is lowered down a 37 degree incline a distance of 5.0m from point

Question

A 4.0kg block is lowered down a 37 degree incline a distance of 5.0m from point A to point B ( the vertical distance 5.0m sin37 degree= 3.0m) A horizontal force (F=10N) is applied to the block between A and B.

Part A)

How much work is done by the force F as the block slides from A to B?

Part B)

How much work is done by gravity as the block slides from A to B?

Part C)

The kinetic energy of the block at A is 10J and at B it is 20J. How much work is done on the block by the force of friction between A and B?

Part D)

What is the coefficient of kinetic friction?

Explanation / Answer

A) work done by the force F = 10Cos(37) *5 = 39.93 N

B) Work done by gravity = 4*9.8*3 = 117.6 N

C) KE (A) = 10J    KE(B) =20J

Total work done = 117.6 + 39.93 = 157.53 J

Work done by frictional force Wf = 157.53 - (20-10) = 147.53 J

frictional force Ff = 147.53/5 = 29.51 N

Normal force = mgCos(37) = 4*9.8*Cos(37) =31.31

co-efficient ofr friction = 29.51/31.31 = 0.62

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