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

Question

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?

Answer 1

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:

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

Answer: i) $F=7.06N;$

ii) $F=8.14N.$