The rear side of a truck is open and a box of mass 20 ~kg is placed on the truck 4 ~m away from the open end. If is 0.15 and g is 10 ~m ~s ⁻² and the truck starts from rest with an acceleration of 2 ~m ~s ⁻² on a straight road, then the box will fall off the truck when it is at a distance of x metre from the starting point. The value of x is

Step-by-step solution

a=2 ~m ~s ⁻², N=m g For the box's motion w.r.t. truck, aligned & m a- N=m a_ B T (m a- m g)=m a_ B T & a_ B T =a- g aligned aligned & =2 ~m ~s ⁻²-(0.15) (10 ~m ~s ⁻² ) & =0.5 ~m ~s ⁻² aligned aligned & s_ B T =u_ B T t+ 1 2 a_ B T t^2 & 4=0+ 1 2 (0.5) t^2 t^2=16 & t=4 ~s . aligned After 4 ~s , the box will fall off the truck. During 4 ~s , the truck travels, x= 1 2 a t^2= 1 2 (2)(4)^2=16 ~m .