A box of mass m is initially at rest on a horizontal surface. A constant horizontal force of mg/2 is applied to the box directed to the right. The coefficient of friction of the surface changes with the distance pushed as `mu=mu_(0)x` where x is the distance fr4om the initial location . For what distance is the box pushed until it comes to rest again?
A
`(2)/(mu_(0))`
B
`(1)/(mu_(0))`
C
`(1)/(2mu_(0))`
D
`(1)/(4mu_(0))`
Text Solution
Verified by Experts
The correct Answer is:
B
Net change in kinetic energy =0 `rArr` net work W=0 `w=intdW=intFdx-intmuNdx=(mg)/(2)xx0mg1_(0)int_(0)^(x)xdx=0rArrx=(1)/(mu_(0))`
