Physics question

Question

A 6.000 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is 𝜇s=0.655 and the coefficient of kinetic friction is 𝜇k=0.205. At time 𝑡=0, a force 𝐹=23.7 N is applied horizontally to the block. State the force of friction applied to the block by the table at times 𝑡=0 and 𝑡>0.Consider the same situation, but this time the external force 𝐹 is 47.8 N. Again, state the force of friction acting on the block at times 𝑡=0 and 𝑡>0.

Answer 1

I will assume g = 9.81 m/s²

Step 1: Draw a free body diagram:

Where F_G is the force of gravity (the weight of the block), F_N is the normal force, F_F is the force of friction, and F_A is the applied force.

F_G = mg = (6.000)(9.81) = 58.86 N

Summing the forces in the y-direction:

∑F_y = 0

F_N - F_G = 0

F_N = F_G

F_N = 58.86 N

Considering the forces in the x-direction:

F_F = μF_N where initially μ = 0.655

F_F = (0.655)(58.86) = 38.55 N

HOWEVER, since F_F is > F_A there will be no motion and if there is no motion, then a = 0. Since a = 0, when we sum the forces in the x-direction, we get:

∑F_x = ma = 0

F_A - F_F = 0

F_A = F_F

F_F = 6.000 N

So F_F at time t = 0 is 6.000 N

And F_F at time t > 0 is 6.000 N

Now, considering F_A = 47.8 N:

At time t = 0 since F_F = μF_N = (0.655)(58.86) = 38.55 N but in this case the applied force (F_A) is greater than the the force of friction (F_F) meaning there will be motion. So:

∑F_x = ma

meaning the block will start to accelerate.

At t = 0, F_F = 38.6 N

But at time t > 0, the coefficient of friction will switch to the kinetic coefficient of friction so F_F = (0.205)(58.86) = 12.1 N