Question

The rear side of truck is open and a box of 40 kg mass is placed 5m away from its open end. 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 2m/s${}^2$. At what distance from the starting point of the truck does the box fall off (ignore the size of the box)?

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Answer 1

The situation is shown in figure. We are given that mass of the box, m = 40 kg

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

Acceleration of the truck, a${}_0$ = 2 m/s${}^2$. Let's go in the frame of the truck

The pseudo force $\left(F_0\right)$ acting on the box due to acceleration $\left(a_0\right)$ of the truck is opposed by the frictional force (f) between the box and the surface below it.

Resultant force acting on the box, i.e., F = $\left(F_0\right)$ - f

Or $F=m{a}_0-\mu R=m{a}_0-\mu mg=m(a_0-\mu g)orF=40(2-0.15\times 10)=20N$

Let a be the acceleration produced in the box relative to the truck. Clearly, a =$\frac{F}{m}=\frac{20}{40}=0.5m/s^2$

Free - body diagram: to simplify, draw the free-body diagram for the box as shown in figure (b). Clearly, R = mg and net force acting on the box, i.e.,

$F=F_0-f=m{a}_0-\mu mgorF=m(a_0-\mu g)$

Acceleration produced in the box relative to the truck, i.e.,$a=\frac{F}{m}=\frac{m(a_0-\mu g)}{m}=a_0-\mu g=2-0.15\times 10=0.5m/s^2$

it is the time taken by the box to fall ,i.e., to cover the distance of 5m,then from

$s=v_0\frac{1}{2}a{t}^2,wegets=\frac{1}{2}\left(0.5\right)t^2(as{v}_0=0,s=5m)or{t}^2=\frac{10}{0.5}$

if s' is the distance covered by the truck during this time (i.e., t)

$s^{\prime }=v_{0}t+\frac{1}{2}a{t}^2=\frac{1}{2}\times 2\times \left(\frac{10}{0.5}\right)=20m(as{v}_0=0,a_0=2m/s^{-2})$