Question

F A 6.100 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is us = 0.605 and the coefficient of kinetic friction is f = 0.155. At time i = 0, a force F = 22.3 N is applied horizontally to the block. State the force of friction applied to the block by the table at times t = 0 and 1 > 0. m Ms,k 1=0 36.204 N 1 > 0 9.275 N Incorrect Incorrect Consider the same situation, but this time the external force F is 44.9 N. Again, state the force of friction acting on the block at times I = 0 and 1 > 0. 1=0 36.204 N 1> 0 9.275 N

Answer 1

Mark ng. Mass of block, ma 6.100 kg coefficient of static friction , Ms = 0.605 and coefficient of kinetic friction, Mus o.155 mg Normal fonce (6.1kg) *(9.81 misa) the block on opposes the applied force is, The maximum static friction fonce that (0.605) X x 59.841 N = 36. 204 N fomax is 22.3 N. is smaller than fmax, so the block How, given that, the applied fonce The applied fonce does not move. Thus, f- f = 0 or, faf 22: 3 N.

is Therefore, the frictional fonce at t-o is 22.3 N and the frictional fonce at tro 22.3 N MON considering applied force, 44.9 M je, the applied force. greater than the friction al fonce Therefore the block will stant to more. Therefore, the toictional fonce at to is , f = fman 36. 204 N. The fictional force at time, tyo is, f = My mg (0.185)* (6.1 xg)*60.84 mis) :9.275 N