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Answer»

velocity = 36km/hr = 36*5/18 = 10m/s

work done by forces = change in K.E

and Finally the car gets stop so ∆K.E = 1/2*(m)*(10)² = 50m

so, work done my gravity = mgh = 10mh

and work done by friction along the inclined path having distance h/cos60° = 2h is = μmgcos(60)*2h= 0.1*m*10*h = mh

no equating all work done=> mh +10mh = 50m=> 11h = 50=> h = 50/11 m



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