# Answer to Question #122162 in Mechanics | Relativity for Shirshak Aryal

Question #122162
A 6kg box is pushed across a flat table by a horizontal force F. If the box is moving at constant velocity of 0.35m/s and coefficient of kinetic friction is 0.12, find F. What is the magnitude of F if the box is moving at constant acceleration of 0.18 m/s^2?
1
2020-06-15T10:26:19-0400

Solution.

"m=6 kg;"

"v=0.35m\/s;"

"\\mu=0.12;"

"a=0.18m\/s^2;"

i)Since the box moves at a constant speed, the equivalent force applied to it is zero. The force of friction is balanced by the force of traction.

"F=\\mu mg;"

"F=0.12\\sdot6kg\\sdot 9.81N\/kg=7.06N;"

ii)If the box moves with acceleration, then the equivalent causes the occurrence of this acceleration.According to Newton's second law we have:

"F-\\mu mg=ma\\implies F=ma+\\mu mg;" "F=6kg\\sdot0.18m\/s^2+0.12\\sdot6kg\\sdot9.81N\/kg=8.14N;"

ii) "F=8.14N."

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!