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.

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Infinity Learn · Accepted Answer

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, $$ma1=$$μ$$R+mgsin$$θ∴   Retardation,  $$a1=$$μ$$R+mgsin30$$°mAlso,  $$R=mgcos30$$°⇒  $$a1=$$μ$$mgcos30$$°$$+mgsin30$$°$$m=$$μ$$g32+g2$$                                                             …(ii)Case$$ $$2$$ For body coming down the plane,$$s=ut+12a2t22As$$     u $$=$$ $$0$$⇒  $$t2=2s/a2$$                                                      …$$(iii)Downward$$ acceleration,$$a2=mgsin30$$°$$-$$μ$$Rm=mgsin30$$°$$-$$μ$$mgcos30$$°$$m=g2-$$μ$$g32$$                                                       …$$(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 getμ$$g32+g2=4g2-$$μ$$g32Solving$$ μ, we getμ$$=35=1.7325$$⇒  μ$$=0.346$$

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.

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