A 200kg block is at rest on a horizontal table. thecoefficient of static and kin

Question

A 200kg block is at rest on a horizontal table. thecoefficient of static and kinetic friction between the table andthe block is 0.60 and 0.40 respectively. An 1800N force is thenapplied on the block. b) will the block start to slide? c) what is the mgnitude and direction of the friction force onthe block? d) what is the acceleration of the block? A 200kg block is at rest on a horizontal table. thecoefficient of static and kinetic friction between the table andthe block is 0.60 and 0.40 respectively. An 1800N force is thenapplied on the block. b) will the block start to slide? c) what is the mgnitude and direction of the friction force onthe block? d) what is the acceleration of the block?

Explanation / Answer

The frictional force would keep the block at rest unlessthe applied force exceeds the limiting force of static friction which is equal to Fs,max = s*N (for smaller applied force, the frictional force is equaland opposite to applied force), In this case, N = normal force = M*g since in the verticaldirection, N balances M*g So, Force needed to move the block = Fapp = 0.60*200.0*9.8 N = 1176 N Now, once the block starts moving, the frictional force becomes constant, and equal to Fk = k*N = k*M*g = 0.40*200.0*9.8 N = 784 N However the applied force remains same. So, writing the force-acceleration equation,
yes the block start to slide Fapp - Fk = M*a => a = [0.60 - 0.40]*9.8 = 1.96 m/s2

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