A 4.240 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 = 0.255. At time t = 0, a force F = 11.6 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 static and kinetic friction.
1. Initially, the block is at rest and the force applied (F = 11.6 N) is less than the maximum static friction force.
- The force of friction is equal to the force applied since the block remains stationary.
- Therefore, the force of friction at t = 0 is equal to 11.6 N.
2. As time progresses and the force applied remains constant, the block starts moving.
- When the block transitions from being stationary to moving, the force of friction changes from static to kinetic friction.
- The force of kinetic friction is given by the equation Fk = μk * N, where μk is the coefficient of kinetic friction and N is the normal force.
To calculate the normal force, we need to consider the weight of the block:
Weight = mass * acceleration due to gravity
Weight = 4.240 kg * 9.8 m/s^2
Weight = 41.552 N
3. The normal force (N) is equal to the weight of the block since it is resting on the table.
- Therefore, N = 41.552 N.
4. Now, we can calculate the force of friction at t > 0 using the equation Fk = μk * N.
- The force of kinetic friction at t > 0 is Fk = 0.255 * 41.552 N.
- Therefore, the force of friction at t > 0 is 10.62676 N, which can be rounded to 10.6 N.
In summary, the force of friction applied to the block by the table is:
- At t = 0: 11.6 N (static friction)
- At t > 0: 10.6 N (kinetic friction)
To determine the force of friction applied to the block by the table at different times, we need to consider whether the block is at rest or in motion.
1. If the block is at rest (static friction):
At t = 0, when the force F = 11.6 N is applied horizontally to the block, we compare this force to the maximum static friction force. The maximum static friction force can be calculated using the formula:
Fs(max) = μs * N
where Fs(max) is the maximum static friction force, μs is the coefficient of static friction, and N is the normal force acting on the block (equal to the weight of the block).
N = m * g
where m is the mass of the block and g is the acceleration due to gravity.
N = 4.240 kg * 9.8 m/s^2 = 41.552 N
Substituting the values into the formula:
Fs(max) = 0.455 * 41.552 N = 18.91976 N (approximately)
Since the applied force of 11.6 N is less than the maximum static friction force, the block remains at rest. Therefore, the force of friction applied to the block by the table is equal to the applied force, which is 11.6 N.
2. If the block is in motion (kinetic friction):
At t > 0, if the applied force F exceeds the maximum static friction force, the block will start moving, and the force of friction applied will change to the kinetic friction force.
The kinetic friction force can be calculated using the formula:
Fk = μk * N
where Fk is the kinetic friction force, and μk is the coefficient of kinetic friction (given as 0.255).
To calculate the normal force N, use the same formula as before:
N = m * g = 4.240 kg * 9.8 m/s^2 = 41.552 N
Substituting the values into the formula:
Fk = 0.255 * 41.552 N = 10.60436 N (approximately)
Therefore, once the block starts moving, the force of friction applied to the block by the table will be approximately 10.60436 N.
before moving: mg* μs
after moving: mg*μk