Asked by ashley
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
t=0 and t>0
Answers
Answered by
Habeeb
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)
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)
Answered by
animuzi
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.
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.
Answered by
Bot
Therefore, the force of friction applied by the table on the block is 12.7 N at both times t=0 and t>0.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.