A 1.920 kg block of wood rests on a steel desk. The coefficient of static fricti
ID: 1352551 • Letter: A
Question
A 1.920 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is s = 0.605 and the coefficient of kinetic friction is k = 0.255. At time t = 0, a force F = 7.01 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 14.1 N. Again state the force of friction acting on the block at the following times: t=0 and t>0
Explanation / Answer
static friction, Fs = mue_s*m*g
= 0.605*1.92*9.8
= 11.4 N
so, when F = 7.01 N
at t = 0,
frictional force = static friction
= 7.01 N
at t > 0 also
frictional force = static friction
= 7.01 N
because F < Fs
so, the body does not move.
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when F = 14.1 N
at t = 0,
frictional force = static friction
= 7.01 N
at t > 0 also
frictional force = kinetic friction
= mue_k*m*g
= 0.255*1.92*9.8
= 4.8 N
here the body will be in motion, beacuse F > Fs
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