Question

A particle is projected up directly up a rough plane of inclination ¥ with velocity u.if after coming to rest the partcle returns to its starting point with a velocity v.show that the coefficient of friction is €=[{u²-v²}÷{u²+v²}tan¥]?

Answer 1

Please see the proof below.

Explanation:

The initial velocity is `=u`

The inclination of the plane is `=theta`

The coefficient of friction is `=mu`

Resolving in the direction up the plane

The acceleration is

`a=-mg(sintheta+mucostheta)`

The equation of motion is

`v^2=u^2+2as`

`0=u^2-2(mg(sintheta+mucostheta))s`

`s=u^2/(2mg(sintheta+mucostheta))`

Resolving in the direction down the plane

The final velocity is `=v`

The acceleration is

`a_1=mg(sintheta-mucostheta)`

Therefore,

`v^2=u_1^2+2a_1s`

`v^2=0+2mg(sintheta-mucostheta)*u^2/(2mg(sintheta+mucostheta))`

`v^2=(sintheta-mucostheta)/(sintheta+mucostheta)u^2`

Dividing by `costheta`

`v^2=(tantheta-mu)/(tantheta+mu)u^2`

`v^2(tantheta+mu)=u^2(tantheta-mu)`

`v^2tantheta+muv^2=u^2tantheta-muu^2`

`muv^2+muu^2=u^2tantheta-v^2tantheta`

`mu(u^2+v^2)=(u^2-v^2)tantheta`

`mu=(u^2-v^2)/(u^2+v^2)tantheta`