A block whose weight is 41.6 N rests on a horizontal table. A horizontal force of 33.3 N is applied to the?

Question

A block whose weight is 41.6 N rests on a horizontal table. A horizontal force of 33.3 N is applied to the block. The coefficients of static and kinetic friction are 0.699 and 0.352, respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration? If the block does not move, give 0 m/s2 as the acceleration.

Answer 1

Okay,
The normal force is equal to the weight of the block and the force of friction is equal to the normal force times the coefficient of friction.

Now if the dfference between the applied force and the frictional force is > zero, the block moves. So

F - us*mg = 33.3 - 0.699*41.6 = 30.4 Nts so block moves

Now block accelerates if F - coef of kin frict*N > 0

So a = (F - uk*mg)/m = (F -uk*W)/(W/g) = 4.4 m/s^2

Answer 2

well, the friction foce is mu_s(coefficient of static friction)*F_n (natural force). In this case, the natural force is 41.6 N, so the static friction is 29.0784. Therefore, the net force on the block is 4.2216 N. Now we can use F = ma, and solve for acceleration, which equals 0.942307692. (note that mass = 41.6 N/9.80 m/s^2, not 41.6 N). After the block is pushed, however, the only horizontal force acting on the block will be the kinetic friction, which equals mu_k*F_n = 14.6432N, so the block will decelerate at (using F=ma again) 3.4496 m/s^2.

I'm pretty sure this is right, but I didn't write stuff down, so...

Wait, is the horizontal force constant or not?

Answer 3

The force of friction is given by:

F = u * N, where u is coefficient of friction, and N is the normal force (in this case N = mg). So the static friction force is

F = 0.699mg = 0.699 (41.6 N) = 29.1 N

This is the static friction force. This is the minimum force we needed to make the block move. So 33.3 N is enough to make the block move.

What is the friction force once the block starts moving? F = umg, in this case, equals to F = (0.352)(41.6 N) = 14.6 N. Therefore, the total force then equals to 33.3 N - 14.6 N = 18.7 N, since friction acts in the opposite direction than which it is pushed.

Thus a = F/m = 18.7N/(41.6N / g) = 4.41 m/s^2

Answer 4

The force required to overcome the static friction is just over 29 Newtons.
The block which weighs 41.6 Newton has a mass of 4.2 Kg
The kinetic friction is 14.6 newtons
The resultant fore is therefore 18.7 Newtons
The acceleration is therefore 4.4 meters per Second per Second