Question

A 2.040 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is μs = 0.605 and the coefficient of kinetic friction is μk = 0.255. At time t = 0, a force F = 7.45 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following times: t=0 and t > 0.

Consider the same situation, but this time the external force F is 15.0 N. Again state the force of friction acting on the block at the following times: t=0 and t > 0.

Answer 1

the normal force is " N "

N = mg

frictional force f_s

f_s = _s N

= _s m g

= 0.605 x 2.04 x 9.8

f_s = 12.09516 N

but the applied force is F = 7.45 N

f_s F

so we can write that f_s = 7.45 N

at t = 0 , 7.45 N

at t > 0 , 7.45 N

now,

F_external = 15 N

f_k = _k m g

= 0.255 x 2.04 x 9.8

f_k = 5.09796 N

t = 0 , f_s = 12.09516 N

t > 0 , f_k = 5.09796 N