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The rear side of a truck is open and a box of 40 kg mass is placed 5 m from the open end The coefficient of friction between the box and the surface below it is 0.15 On a straight road the truck starts from rest and accelerates with `2 m//s^(2)` At what distance from the starting point does the box fall off the truck ? Ignore the size of the box . |
Answer» Here , mass of box , ` m = 40 kg ` acceleration of truck ,`a = 2 m//s^(2)` distance of box from open end , s = 5 m coefficient of friction ` mu = 0.15` Force on the box due to accelerated motion of truck `F = ma = 40 xx 2 = 80 N` This force is in forward direction Reaction F of this force on box = 80 N in backward dirction This reaction is opposed by force of limiting friction `f = mu R = mu mg = 0.15 xx 40 xx 9.8 = 58.8` N in the forward direction `:.` Net force on the box in backward direction `p = F - f = 80 - 58 .8 = 21 .2` N backwards acceleration of block `a = (p)/(m)` ` a = (21.2)/(40) = 0 .53 m//s^(2)` If t is time taken by the box to travel s = 5 m and fall off the truck then from ` s = ut + (1)/(2) at^(2)` `5 = 0 + (1)/(2) xx 0.53 t^(2)` ` t = sqrt((2 xx 5 )/(0.53 ))= 4.34 s` If the truck travels a distance x during this time then from `s = ut + (1)/(2) at^(2)` `x = 0 + (1)/(2) xx 2 (4.34)^(2) = 18.84 m` . |
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