Question

The figure shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angle θ = 20°. What is the magnitude of the acceleration of the block across the floor if (a) μs = 0.640 and μk = 0.500 and (b) μs = 0.430 and μk = 0.300?

Answer 1

Horizontal force exerted by the block which will be given as -

F_x = F cos

F_x = (0.5 mg) cos 20⁰

F_x = 0.4698 mg

Normal force exerted by the block which will be given as -

N = mg - F sin

N = mg - (0.5 mg) sin 20⁰

N = mg (1 - 0.171)

N = 0.829 mg

What is the magnitude of an acceleration of the block across the floor, if (a) _s = 0.640 and _k = 0.500.

Static frictional force exerted by the block which will be given as -

f_s = _s N

f_s = (0.640) (0.829 mg)

f_s = 0.5305 mg

We know that, F_x < f_s.

The static friction force is greater than the applied force. So, the block will not move without more force or less friction.

Then, the acceleration of the block will be zero.

a = 0 m/s²

What is the magnitude of an acceleration of the block across the floor, if (a) _s = 0.430 and _k = 0.300.

Static frictional force exerted by the block which will be given as -

f_s = _s N

f_s = (0.430) (0.829 mg)

f_s = 0.3564 mg

Kinetic frictional force exerted by the block which will be given as -

f_k = _k N

f_k = (0.300) (0.829 mg)

f_k = 0.2487 mg

We know that, F_x > f_s.

The static friction force is less than the applied force. So, the block will move.

From Newton's 2^nd law of motion, we get

a = F_net / m

a = (F_x - f_k) / m

a = [(0.4698 mg) - (0.2487 mg)] / m

a = (0.2211) mg / m

a = [(0.2211) (9.8 m/s²)]

a = 2.16 m/s²