A body of mass 5 X 10-3 kg is launched upon a rough inclined plane making an angle of 30° with the horizontal. Obtain the coefficient of friction between the body and the plane, if the time of ascent is half of the time of descent.

According to question, the situation can be shown asCase 1 For body projected up the plane,   v=0Using third equation of motion,  v2=u2-2a1s⇒  0=u2-2a1s∴  u=2a1sAlso    v=u-a1t1⇒  0=u-a1t1⇒  t1=u/a1∴  t1=2a1sa1=2sa1                                                   …(i) As, ma1=μR+mgsinθ∴   Retardation,  a1=μR+mgsin30°mAlso,  R=mgcos30°⇒  a1=μmgcos30°+mgsin30°m=μg32+g2                                                             …(ii)Case 2 For body coming down the plane,s=ut+12a2t22As     u = 0⇒  t2=2s/a2                                                      …(iii)Downward acceleration,a2=mgsin30°-μRm=mgsin30°-μmgcos30°m=g2-μg32                                                       …(iv)Given,     t1=t2/2Substituting values of t1 and t2 from Eqs. (i) and (iii), we get2sa1=122sa2Now, squaring both sides, we geta1=4a2                                                                …(v)Substituting values of a1 and a2 from Eqs. (ii) and (iv) in Eq. (v), we getμg32+g2=4g2-μg32Solving μ, we getμ=35=1.7325⇒  μ=0.346