The figure shows a block of mass m resting on a 20° slope.The coefficients of friction between the block and the surface of the slope are $\mu _ { \mathrm { S } } = 0.64$ and $\mu _ { \mathrm { k } } = 0.54$ It is connected using a very light string over an ideal pulley to a hanging block of mass 2.0 kg.The string above the slope pulls parallel to the surface.What is the minimum mass m so the system will remain at rest when it is released from rest?
A)2.1 kg
B)3.3 kg
C)1.3 kg
D)3.6 kg

Question

The figure shows a block of mass m resting on a 20° slope.The coefficients of friction between the block and the surface of the slope are $\mu _ { \mathrm { S } } = 0.64$ and $\mu _ { \mathrm { k } } = 0.54$ It is connected using a very light string over an ideal pulley to a hanging block of mass 2.0 kg.The string above the slope pulls parallel to the surface.What is the minimum mass m so the system will remain at rest when it is released from rest? 
A)2.1 kg 
B)3.3 kg 
C)1.3 kg 
D)3.6 kg

Quizplus · Accepted Answer

To find the minimum mass, we need to consider the maximum force of friction that can act on the block.

The weight of the hanging block is 2.0 kg * 9.81 m/s^2 = 19.62 N.

The force that the string exerts on the block is also 19.62 N (because the system is at rest).

This force is parallel to the slope, so we need to find the component of the weight that is parallel to the slope:

19.62 N * sin 20° = 6.70 N

The maximum force of friction is $\mu _ { \mathrm { S } } m g \cos \theta$ = 0.64 * m * 9.81 m/s^2 * cos 20° = 6.28 m N.

Since the system is at rest, the force of friction must be equal to the force parallel to the slope:

6.28 m N = 6.70 N

Solving for m, we get:

m = 2.1 kg

The Figure Shows a Block of Mass M Resting on a 20°