A 6.200 kg block of wood rests on a steel desk. The coefficient of static fricti
ID: 1330836 • Letter: A
Question
A 6.200 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is s = 0.455 and the coefficient of kinetic friction is k = 0.305. At time t = 0, a force F = 17.0 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 ???? N
T>0 ???? N
Consider the same situation, but this time the external force F is 34.3 N. Again state the force of friction acting on the block at the following times:
T=0 ???? N
T>0 ???? N
Explanation / Answer
Given
Mass m= 6.2 kg
Coefficient of static friction µs = 0.455
Coefficient of kinetic friction µk = 0.305
Applied force = 17.0 N
Solution
A)
At time = 0 s
The block is at rest due to the static friction
Fs = µsFnormal
= 0.455 x mg
= 0.455 x 6.2 x 9.8
= 27.6458 N
A minimum force of 27. 64 N is required to move the block since the applied force here is lesser than the required for the block will continue to be at rest. Which means the samo amount of static friction will be present at t > 0. To summarize
At t = 0 ; Fs = 27.64 N
At t > 0 ; Fs = 27.64 N
B)
Here applied force F = 34.3 N
Here F>Fs so the block will move. When an object is in motion, instead of static friction a different kind of friction called kinetic friction will be present
Fk = µkFnormal
= 0.305 * 6.2*9.8
= 18.5318 N
To summarize
At t = 0 ; Fs = 27.64 N
At t > 0 ; Fk = 18.53 N