A block of mass m=1.97 kg is stacked on top of a block of massM=3.06 kg and the

Question

A block of mass m=1.97 kg is stacked on top of a block of massM=3.06 kg and the two move together on a horizontal surface with aconstant acceleration of magnitude a=3.03 m/s2 pointingto the right under the influence of an external horizontal force ofmagnitude F0=39.6 N which is applied to mass M (seefigure). The coefficient of static friction between mass m and massM is s=0.670. The coefficient of kinetic frictionbetween block m and block M is not given. Nor is the coefficient ofkinetic friction given between block M and the horizontalsurface.
a) What is the magnitude of the force of friction exerted by blockM on m?
c) What is the magnitude of the force of friction exerted by thehorizontal surface on block M?

Explanation / Answer

a)the magnitude of the force of friction exerted by block M onm is fk= s* n the force acting on mass m in y direction is Fy = n + (-w) ------------(1) the force acting on mass M in y direction is Fy = n1 + (-w1) = 0 or n1 = w1 -------------(2) from (1) and (2) n + (-w) = w1 or n = w1 + w or fk= s* (w1 + w) =s* (M * g + m * g) = s* g *(M + m) g = 9.8 m/s^2 b)the magnitude of the force of friction exerted by thehorizontal surface on block M is Fo - fk= (m + M) * a or fk= Fo - (m + M) * a the force acting on mass M in y direction is Fy = n1 + (-w1) = 0 or n1 = w1 -------------(2) from (1) and (2) n + (-w) = w1 or n = w1 + w or fk= s* (w1 + w) =s* (M * g + m * g) = s* g *(M + m) g = 9.8 m/s^2 b)the magnitude of the force of friction exerted by thehorizontal surface on block M is Fo - fk= (m + M) * a or fk= Fo - (m + M) * a

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