Question

The rear side of a truck is open and a box of mass 40 kg is placed 5 m away from the open end. The coefficient of friction between the box and the surface below it is 0.15. The truck starts form rest with an acceleration of $2ms^{- 2}$on a straight road. At what distance from the starting point does the box fall off the truck?

• A. 20 m
• B. 30 m
• C. 40 m
• D. 50 m

Answer 1

Answer: (A)

20 m

Answer 2

$a = 2 \mathrm {~ms} ^ { - 2 } / 7 \mathrm {~m}$

Here,

Mass of the box M =40kg

Acceleration of the truck, $a = 2ms^{- 2}$

Distance of the box from the rear end, $d = 5m$

Coefficient of friction between the box and the surface below it $\mu = 0.15$

The various forces acting on the block are as shown in the figure

As the trunk moves in forward direction with the acceleration $a = 2ms^{- 2}$the box experiences a forces F in

$F = Ma = (40kg)(2ms^{- 2}) = 80N$in backward direction

Under the action of this force, the box will tend to move toward the rear end of the trunk. As it does so, its motion will be opposed by the force of friction which acts in the forward direction and it is given by

$f = \mu N = \mu Mg = 0.15 \times 40 \times 10 = 60N$

The acceleration of the box relative to trunk toward the rear end is

$a_{1} = \frac{F - f}{M} = \frac{80N - 60N}{40kg} = 0.5ms^{- 2}$

Let t be the time taken for the box to fall off the trunk

$d = 0 \times t + \frac{1}{2}a_{1}t^{2}(\therefore u = 0)$

$5 = \frac{1}{2} \times 0.5 \times t^{2}$

$t = \sqrt{\frac{2 \times 5}{0.5}} = \sqrt{20}$

During this time, the truck covers a distance x

Using $S = ut + \frac{1}{2}at^{2}$

We get

$x = 0 \times t + \frac{1}{2} \times (\sqrt{20})^{2}\therefore u = 0$

x = 20m