A 4.170 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is μs = 0.505 and the coefficient of kinetic friction is μk = 0.255. At time t = 0, a force F = 12.7 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

Well, well, well, looks like we've got a physics problem on our hands! Don't worry, I'm here to clown around and help you out.

At time t=0, when the force of 12.7 N is applied, the block is not moving yet. This means we're dealing with static friction. So, the force of friction applied to the block by the table is equal in magnitude but opposite in direction to the applied force. We can calculate it using the formula:

Friction(force) = coefficient of static friction × normal force

The normal force is the weight of the block, which we can calculate by multiplying the mass of the block by the acceleration due to gravity (g ≈ 9.8 m/s²).

Normal force = mass × gravity
Normal force = 4.170 kg × 9.8 m/s²

Now, let's plug in the values and do some math:

Friction(force) = 0.505 × (4.170 kg × 9.8 m/s²)

Calculate that out, and you'll find the force of friction applied to the block by the table at t=0.

Now, for t > 0, things get a little different. Once the block starts moving, we switch from static friction to kinetic friction. The force of kinetic friction is given by:

Friction(force) = coefficient of kinetic friction × normal force

Again, we'll calculate the normal force as before, and then plug in the values:

Friction(force) = 0.255 × (4.170 kg × 9.8 m/s²)

And there you have it! Now you know the forces of friction applied to the block by the table at t=0 and t>0.

In both cases, the force of friction is 12.7 N because the box won't move. It's just being pushed once and halts. Hope this helps:

𝑓≤𝜇s𝑁
This means that the force of static friction can take on any value from 0
to 𝜇𝑁. Normal force = mg

static force vector=[0,𝜇s𝑚𝑔]
= [0,0.505×(4.17 kg)×(9.81 m/s2)] = [0,20.7]

At time 𝑡=0 the applied force is 𝐹=12.7 N, well below the maximum static frictional force which is 20.7 N. Therefore, the frictional force can and will rise to match this external force. Since the object remains motionless, the frictional force can maintain this value indefinitely, that is, for 𝑡>0.

Therefore, the force of friction applied by the table on the block is 12.7 N at both times t=0 and t>0.

To determine the force of friction applied to the block by the table at different times, we need to consider the conditions of static and kinetic friction.

1. At t = 0:
When the force F = 12.7 N is applied horizontally to the block, we need to determine whether the block will start moving or remain stationary. To do this, we compare the applied force with the maximum static friction.

The maximum static friction force (fs) can be calculated using the formula:
fs = μs * N
where μs is the coefficient of static friction and N is the normal force acting on the block.

The normal force (N) is the force exerted by the desk on the block, which equals the weight of the block (mg):
N = m * g

Given:
m = 4.170 kg (mass of the block)
g = 9.8 m/s^2 (acceleration due to gravity)
μs = 0.505 (coefficient of static friction)

Calculations:
N = m * g
N = 4.170 kg * 9.8 m/s^2
N ≈ 40.866 N

fs = μs * N
fs = 0.505 * 40.866 N
fs ≈ 20.623 N

Since the applied force F (12.7 N) is less than the maximum static friction force (20.623 N), the block remains stationary. Therefore, the force of friction applied by the table at t = 0 is equal to the applied force: 12.7 N.

2. At t > 0:
Once the block starts moving, the force of friction transitions from static friction to kinetic friction. The force of kinetic friction (fk) is given by:
fk = μk * N
where μk is the coefficient of kinetic friction and N is the normal force.

Calculations:
fk = μk * N
fk = 0.255 * 40.866 N
fk ≈ 10.425 N

Once the block starts moving, the force of friction is equal to the force of kinetic friction, which will be approximately 10.425 N at t > 0.

Friction = uN

N=mg so

F=umg

F at t(0) = (.505)(4.17 kg)(9.8 m/s^2)

F at t>0 = (.255)(4.17 kg)(9.8 m/s^2)