Question

1)When the three blocks below are released from rest, they accelerate with a magnitude of 0.500m/s2. Block 1 has mass M, block 2 has 2M, and block 3 has 2M. What is the coefficient of kinetic friction

2)A block of mass m is held stationary on a ramp by the frictional force on it from the ramp. A force ?,directed up the ramp, is then applied to the block and gradually increased in magnitude from zero. During the increase, what happens to the direction and magnitude of the frictional force on the block?

3)The figure below shows an initially stationary block of mass m = 2.00 kg on a floor. A force of magnitude 10.0 N is then applied at upward angle θ = 20°. What is the magnitude of the acceleration of the block across the floor if (a) μs = 0.600 and μk = 0.500 and (b) μs = 0.400 and μk = 0.300?

Answer 1

Dear student, my answers to the 1st & 2nd questions are to be referred. Kindly avoid solely depending on the answer to my 3rd question.

Here are the answers to each question.

1) The coefficient of Kinetic Friction (_k) can be calculated by the equation ,

_k = (5*a - g) / (- 2* g) = (5*.500-9.8) / (-2*9.8) = 0.372 ( where a is the given acceleration and g is the acceleration due to gravity which is equal to 9.8 m/s²).

2) By Newton's law, when force we apply increases, the frictional force also increases until it reaches a certain maximum value. When the applied force is greater than the maximum value of frictional force, the object will move.

Here in the question, when we apply force to the block of mass m, the magnitude of frictional force also increases until a certain maximum value of frictional force and when the magnitude of applied force becomes more than the maximum value of frictional force, the block will start to move.

The direction of frictional force will always be opposite to that of the applied force.

3) In the question it is given that,

The magnitude of force, F = 10 N.

By Newton's second law of motion, since there is no vertical motion, the total force in this direction is zero.

Therefore, we can write that,

0 = N + F * sin 20- mg

, N = mg-F*sin20 = mg (1- 10 * sin20) = -2.42 mg

a) The maximum static friction is given by, F_s = _s * N = 0.600 * - 2.42 mg = -1.452 mg

The maximum kinetic friction is given by, F_k = _k * N = 0.500 * -2.42 mg = -1.21 mg

The component of applied force in the horizontal direction = F * cos20 = 10 * 0.939 = 9.39 mg

The net force on the block is F_net = F * cos20 - F_k = 9.39 mg - (-1.21mg) = 10.6 mg

Therefore, by Newton's second law, the net acceleration , a = (F_net / m) = 10.6 mg / m = 10.6 g

b) The maximum static friction is given by, F_s = _s * N = 0.400 * - 2.42 mg = -0.968 mg

The maximum kinetic friction is given by, F_k = _k * N = 0.300 * -2.42 mg = -0.726 mg

The net force on the block is F_net = F * cos20 - F_k = 9.39 mg - (-0.726 mg) = 10.11 mg

Therefore, by Newton's second law, the net acceleration , a = (F_net / m) = 10.6 mg / m = 10.11 g