Question

Solve the question ?

Answer 1

(i) Speed of particle as it reaches `B` is given by the kinematic expression

`v^2-u^2=2as`

Inserting given values we get

`v_B^2-3^2=2(2.5)(8)`
`=>v_B^2=40+9`
`=>v_B=7\ ms^-1`

(ii) Let `theta` be angle of incline. And `F_f` be force of friction
As the particle slides down the inclined plane the downwards force along the incline `=mgsintheta`
Downwards acceleration `=2.5\ ms^-2`
Net force `2.5xx0.8=0.8xx9.81xxsintheta-F_f`

`=>F_f=0.8xx9.81xxsintheta-2`

Now work done against force of friction `W=vecF_f*vecs`
Angle between the force of friction and displacement is `=0^@` `=>cos 0^@=1`. Calculating `W` and equating with the given value we get

`(7.848sintheta-2)xx8=7`
`=>sintheta=(7/8+2)/7.848`
`=>theta=21.5^@`