In the figure,the coefficient of kinetic friction between the surface and the larger block is 0.20,and the coefficient of kinetic friction between the surface and the smaller block is 0.30.If F = 14 N and M = 1.0 kg,what is the magnitude of the acceleration of either block? A)2.0 m/s² B)1.3 m/s² C)1.5 m/s² D)1.8 m/s² E)3.5 m/s²

Question

Quizplus · Accepted Answer

To find the acceleration of either block, we need to calculate the net force acting on each block:

For the larger block:
F_friction = coefficient of kinetic friction * normal force
F_friction = 0.20 * (Mg) = 0.20 * (1.0 kg * 9.8 m/s^2) = 1.96 N
Net force = F - F_friction = 14 N - 1.96 N = 12.04 N

For the smaller block:
F_friction = coefficient of kinetic friction * normal force
F_friction = 0.30 * (mg) = 0.30 * (0.5 kg * 9.8 m/s^2) = 1.47 N
Net force = F_friction = 14 N - 1.47 N = 12.53 N

Now, we can calculate the acceleration of each block using Newton's second law:
a = net force / mass

For the larger block:
a = 12.04 N / 1.0 kg = 12.04 m/s^2

For the smaller block:
a = 12.53 N / 0.5 kg = 25.06 m/s^2

Since the two blocks are connected by a rope, they must have the same acceleration. Therefore, the acceleration of either block is the smaller value, which is approximately 1.3 m/s^2.

In the Figure,the Coefficient of Kinetic Friction Between the Surface