Figure shows a block of mass m resting on a 20 slope. The block has coefficients

Question

Figure shows a block of mass m resting on a 20 slope. The block has coefficients of friction s = 0.81 and k = 0.54 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg.

What is the minimum mass m that will stick and not slip?

Express your answer to three significant figures and include the appropriate units

If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

When the block M of mass M1 is not slipping down or up in when the maximum limiting force acts on it. Under this condition the net force acting on the block M as well as hanging block must be zero, so we have, if T is the tension in the string
T = 2×9.8 = 19.6 N, and
Let us first assume that M1 is such that if it is increased a little more it will move down, in this case frictional force will act in the direction of Tension
T+M1×9.8×(cos20)*0.81 - M1×9.8×sin(20) = 0
19.6 + M1× [7.459 - 3.351] = 0.It is obvious that this equation will give us negative value for M1. Hence we have to rewrite the equation considering friction acts in the direction of gravity and stops M from going up
T-M1×9.8(cos20)×0.81 - M1×9.8×sin(20) = 0 or
T - M1[7.459+3.351] = 0 or M1 = 19.6/10.81 = 1.81 kg
So M1 = 1.81 kg. But to call it as minimum is difficult to understand If mass is more than this the system will start moving down but to evaluate M1 we will have to use earlier equation with k=0.54. Using that equation we have

T+M1×9.8×(cos20)×0.54- M1×9.8×sin(20) = 0
T +M1[2.159 - 8.946] = 0 or M1 = 19.6/6.787 =2.852 kg

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