Please show how to get the answers above with steps shown. Block A (mass 5.00 kg

Question

Please show how to get the answers above with steps shown.

Block A (mass 5.00 kg) is on a long incline that is at 53.1 degree above the horizontal. A light rope is attached to block A. The rope passes over a light frictionless pulley and block B (mass 8.00 kg) is suspended from the free end of the rope, as shown in the sketch. The blocks are released from rest and the tension in the rope is 61.5 N as block A moves up the incline. a) What vertical distance does block B move downward in the first 4.00 s after it is released from rest? b) What is the coefficient of kinetic between block A and the surface of the incline?

Explanation / Answer

We apply Newton's second law to each block separately.

For block B we choose downward as positive, while for block A we choose our x axis along the incline with the positive sense upward.

This choice allows us to use the same symbol, a, for the acceleration of each block.

Then for block B, WB - T =mBa,

where mB = 8 kg

WB = 78.4 N

a is acceleration

T is the tension in the rope.

Since the pulley is frictionless, the same tension T will exist on both sides of the pulley.

Then for block A, T - WA sin 53.1° =mAa,

where mA = 5 kg and WA = 49 N.

We can eliminate the tension T by adding the two equations, which yields WB - WA sin 53.1° = (mA + mB)*a. Substituting in the known values, then

78.4 - 49*0.7996 = (5+8)*a

a = 3.01 m/s2

The equation for fall from rest is y = VOyt + 1/2*ayt2,

with VOy = 0. Substituting in ay = 3.01 m/s2 and t=4 s,

we get y = 24.08 m.

b)

If there is a coefficient of kinetic friction then the equation will be

So using Newton's Second Law, you can set up the equation:
F = T - (k mg cos + mg sin ) = ma
T - k mg cos - mg sin = ma
T - ma - mg sin = k mg cos

61.5 - 49 - 49*0.7996 = k *49*0.625

-26.68/30.63 = 0.87

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