A rear side of a truck is open and a box of mass $20 k g$ is placed on the truck 4 metres away from the open end. $μ = 0.15$ and $g = 10 m / s^{2}$ . The truck starts from rest with an acceleration of $2 m / s^{2}$ on a straight road. The box will fall off the truck when it is at a distance from the starting point equal to

Question

Q. The rear side of a truck is open and a box of mass

2

kg is placed on the truck

8

m away from the open end. The truck starts from rest with an acceleration of

2 m / s^{2}

on a straight road. The box will fall off the truck when it is at distance from the starting point equal to(Given,

μ = 0.1

and

g = 10 m / s^{2}

).

Answer 1

Frictional force

= μ m g

0.15 \times 20 \times 10 = 30 N

a = \frac{F}{m} = \frac{10}{20} = 0.5 m / s^{2}

Now,

s = \frac{1}{2} {a t}^{2}

∴ 4 = \frac{1}{2} \times 0.5 {a t}^{2} \Rightarrow t = 4 s e c

Now distance of truck from initial position after t = 4 sec

s = \frac{1}{2} a_{1} t^{2} = \frac{1}{2} \times 2 \times 16 = 16 m

A rear side of a truck is open and a box of mass 20kg is placed on the truck 4 metres away from the open end. μ=0.15 and g=10m/s2 . The truck starts from rest with an acceleration of 2m/s2 on a straight road. The box will fall off the truck when it is at a distance from the starting point equal to