Question

The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 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 1

Solution-

Here, net force acting on the box is given by,

Net Force = Force due to speed of truck - Frictional force

Fnet = Fx - Ff

Fnet = ma - μmg

Fnet = 40×2 - 0.15×40×10

Fnet = 80 - 60

Fnet = 20 N

Net acceleration = Fnet/m

a' = 20/40 = 0.5 m/s^2

Time required for box to fall off-

s = ut + 1/2 a't^2

5 = 0×t + 0.5×0.5×t^2

5 = 0.25t^2

t = √20 s

Distance travelled by truck in √20 s -

s' = ut + 1/2 at^2

s' = 0×√20 + 0.5×2×(√20)^2

s' = 20 m

Truck will travel 20 m before box falling off.