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 ms². At what distance from the starting point does the box fail off the truck? (ignore the size of the box).

Answer 1

Force experienced by box F = ma = 40 × 2 = 80 N frictional force F_friction = µ mg = 0.15 × 40 × 10 = 60 N

∴ Net force on the box = F – F_friction = 80 – 60 = 20 N .

∴ The backward acceleration experienced by box is given by,

a = \(\frac{Net force}{mass}\) = \(\frac{20}{40}\)= 0.5 m/s²

Let‘t’ be the time taken by box to move through 5m backwards We have, 

S = ut + \(\frac {1}{2}\)at²

∴ 5 = 0 × t + \(\frac {1}{2}\) × 0.5 × t²

t = \(\sqrt20\) ≈ 4.47 s

The distance travelled by truck in t = 4.47s is

s = ut +\(\frac {1}{2}\)at² (a = 2m /s²)

s = 0 ×\((\sqrt20)\)\(\frac {1}{2}\) × 2 \((\sqrt20)^2\)

s = 20 m

The box will off the truck after 20 m from starting point.