Solve the question ?

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1 Answer
May 10, 2018

(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^@