Sapling Learning A 6.280 kg block of wood rests on a steel desk. The coefficient

Question

Sapling Learning A 6.280 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is ,0655 and the coefficient of kinetic friction is 0.255. At time t-o, a force F= 24.8 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following s.k times t=0 t>0 Number Number Consider the same situation, but this time the external force F is 50.0 N. Again state the force of friction acting on the block at the following times: t 0 to o Number Number

Explanation / Answer

max friction force applicable = static friction = mg *us = 6.28 *9.8 * 0.655 = 40.311 N

FORCE APPLIED = 24.8 N

BUT SINCE FORCE APPLIED FROM OUTSIDE IS LESS THAN THE MAX FRICTION FORC POSSIBLE HENCE FRICTION WILL BE JUST ENOUGH TO BALANCE THE EXTERNAL FORCE

SO FRICTION AT T= 0 AND T> 0 WILL BE BOTH EQUAL TO 24.8 NEWTONS ONLY

in the second case :

now since the external is greater than friction force max

so at t = 0 frictionforce = 40.311 N

AT t > 0

frictin force = kinetic friction

=mg*uk =   6.28 *9.8 * 0.255 = 15.693 N

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