Question

A 2.360 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is ?_s=0.455 and the coefficient of kinetic friction is ?_k=0.255. At time ?=0, a force ?=6.48 N is applied horizontally to the block. State the force of friction applied to the block by the table at times ?=0 and ?>0.

t = 0 _________________ N

t > 0 __________________ N

Consider the same situation, but this time the external force ? is 13.1 N. Again, state the force of friction acting on the block at times ?=0t=0 and ?>0.

t = 0 _________________ N

t > 0 __________________ N

Answer 1

Given,

m = 2.36 kg ; us = 0.455 ; uk = 0.255 ; F = 6.48 N

t = 0

so at t = 0, static frictional force applies

Fs = us N = us mg

Fs = 0.455 x 2.36 x 9.8 = 10.53 N

Fs = 10.53 N

applied force is lesser than Fs, it will not move

f - F = 0 => F = f

so at t = 0, F = 6.48 N

t > 0 ; F = 6.48 N

Now the applied force is F = 13.1 N which is greater than Fs,

we already calculated, Fs = 10.53 N

so at t = 0 ; F = 10.53 N

at t >0

F = uk mg

F = 0.255 x 2.36 x 9.81 = 5.9 N

at t > 0 ; F = 5.9 N