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**

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