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. 5.22. 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).

Given:

Mass of the block, M = 40 Kg

Co-efficient of friction, � = 0.15

Initial velocity, u = 0ms^-1

Acceleration, a = 2ms^-2

Distance of Box from the end of truck, s = 5m

Using the Newton’s 3^rd law of motion, the force on the Box, F can be written as,

F = M× a

F = 40Kg × 2 ms^-2

F = 80 N

Since there is friction between the truck and the body, we can write frictional force as,

f_s = �× M× g

f_s = 0.15× 40Kg × 10ms^-2

f_s = 60 N.

Since, force of friction is less than that of exerted by the truck, there exists a net force on Box.

F_Net = F-f_s

F_Net = 80 N – 60 N

F_Net = 20 N

This force acts in backward direction.

We can find the acceleration of the Box using Newton’s second Law,

F_Net = M× a

a = 20 N/ 40 Kg

a = 0.5 ms^-2

The box accelerates in backward direction at the rate of 0.5 ms ^-2.

Since, box is 5 m away from end, i.e. 5 m away from falling down, so we can use the second equation of motion to find the time of fall.

Second equation of motion is given as,

s = ut + 1/2 at² …1

Where,

s = distance travelled

u = initial velocity

t = time

a = acceleration

Since box was at rest initially so u = 0ms^-1

Equation 1 can be re-written as,

S = 1/2 at²

→ t = (2s/a)^1/2

→ t = 20^1/2 s

Now, the distance travelled by truck till the time of fall can also be calculated by equation 1

s = ut + 1/2 at²

s = 0 + 0.5 × 2× (20^1/2)²

s = 20 m