A 4.500 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is μs = 0.455 and the coefficient of kinetic friction is μk = 0.155. At time t = 0, a force F = 12.4 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following times
To determine the force of friction applied to the block by the table at different times, we need to consider the condition of static friction and kinetic friction.
1. Initially (t = 0):
When the force of 12.4 N is applied to the block, the block is at rest. So, the static friction is acting on the block. The force of static friction can be calculated using the formula: Fs = μs * N, where Fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force acting on the block.
To find the normal force, we need to balance the forces acting on the block vertically. Since the block is on a desk, the force of gravity (weight) acting downwards is balanced by the support force (N) acting upwards. The weight can be calculated using the formula: weight = mass * gravitational acceleration (g).
Given:
mass (m) = 4.500 kg
gravitational acceleration (g) ≈ 9.8 m/s^2
weight = m * g
weight = 4.500 kg * 9.8 m/s^2 ≈ 44.1 N
Now, the normal force (N) is equal to the weight of the block, which is 44.1 N.
Substituting the values into the formula:
Fs = 0.455 * 44.1 N ≈ 20.07 N
Therefore, at t = 0, the force of friction applied to the block by the table is approximately 20.07 N.
2. When the block starts moving (t > 0):
After a certain time, when the force applied exceeds the maximum force of static friction (Fs), the block starts moving. When the block is in motion, the force of kinetic friction is applied instead of static friction.
To calculate the force of kinetic friction (Fk), we use the formula: Fk = μk * N, where Fk is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force.
Since the normal force remains the same as before (N = 44.1 N), we can substitute the given values into the formula:
Fk = 0.155 * 44.1 N ≈ 6.84 N
Therefore, when the block is in motion (t > 0), the force of friction applied to the block by the table is approximately 6.84 N.