The coefficient of kinetic friction between the surface and the larger block is 0.25,and the coefficient of kinetic friction between the surface and the smaller block is 0.40.If F = 22N and M = 1.0 kg in the figure,what is the magnitude of the acceleration of either block? A)1.8 m/s² B)2.6 m/s² C)1.4 m/s² D)2.2 m/s² E)3.7 m/s²

Question

Quizplus · Accepted Answer

The force acting on the system is F = 22N. 
The larger block experiences a kinetic friction force given by: f1 = μ1* N1 = 0.25 * (Mg - m*g), where N1 = Mg - m*g is the normal force acting on the larger block. 
Similarly, the smaller block experiences a kinetic friction force given by: f2 = μ2 * N2 = 0.4 * (mg), where N2 = mg is the normal force acting on the smaller block. 
The net force acting on the system along the x-direction is given by: 
Fnet = F - f1 - f2 = Ma
where a is the acceleration of the system. 
Substituting the values, we get: 
22 - 0.25*(1*9.8 - 0.2*9.8) - 0.4*(0.2*9.8) = 1*a 
a = 1.4 m/s^2 
Therefore, the magnitude of acceleration of either block is 1.4 m/s^2.

The Coefficient of Kinetic Friction Between the Surface and the Larger