Question

A body is released from rest from top of an inclined plane of inclination with horizontal
Coefficient of kinetic friction varies with distance s from the top according to relation
u= ks. After what time, will the body stop and what will be the distance travelled by the
body in this time?​

Answer 1

Answer:

Explanation:

The net downward force on the body at a distance x is

f(x)=mgsinθ−μmgcosθ

=mg(sinθ−μcosθ)

=mg(sinθ−kxcosθ)

∴f(x)=0 at a value of x=xo given by

sinθ−kxocosθ=0

Which gives  xo=tanθk

it will stop at tanΘk

and time taken is ..

at x=x1 we will have maximum velocity and after that we will again get deceleration x1=tanθ/k

till x=x1 time taken is g(sinθ-kxcosθ)=dv/dt*(dx/dx)

dx/dt is v

now cross multiply and integrate using appropriate limits

we get v=√(gx1sinθ-kx1²cosθ/2)

using this use v=u+at

and use the same method to find final time and add

please mark it as brainliest cause i have taken a lot of efforts