## Sliding Box

#### Problem

A 10kg box is placed on the back of a lorry, and as the lorry accelerates uniformly from 0 to 36 kmh^{-1} in 5 seconds, the box begins to slide. The lorry then continues to drive at a constant speed and the box comes to rest.

If the box slides 2 metres in total, and assuming `g`=9.8 ms^{-2}, find μ, the coefficient of friction between the box and the surface.

#### Solution

If truck accelerates from 0 to 36 kmh^{-1}=10 ms^{-1} in 5 seconds, acceleration is 2 ms^{-2}.

Consider the diagram.

The limiting equilibrium, R=98μ.

For simplicity, we shall assume throughout that the coefficient of friction remains constant and that the dynamic friction (once the box begins sliding) is the same as the static (limiting) friction.

As truck accelerates, F = `ma` = 102 = 20N, and as we are told that the box slides, we know that F must exceed R and the box will accelerate.

F R = `ma`: 20 98μ = 10`a`_{1}, `a`_{1} = 2 9.8μ`s`=`ut`+½`at`^{2}: `s`_{1}=0+½(2 9.8μ)5^{2}=(25/2)(2 9.8μ)`v`=`u`+`at`: `v`=5(2 9.8μ)

When truck reaches 36 kmh^{-1} there is no accelerating force acting on the box by the motion of the truck, so F = 0 and R will act against the motion of the box, bringing it to rest.

F R = `ma`: 098μ=10`a`_{2}, `a`_{2}=-9.8μ`v`^{2}=`u`^{2}+2`as`: 0=(5(2 9.8μ))^{2}19.6μ`s`_{2}, `s`_{2}=25(2 9.8μ)^{2}/(19.6μ)

Therefore, s_{1}+`s`_{2}=(25/2)(2 9.8μ)+25(2 9.8μ)^{2}/(19.6μ)=2.

Multiply through by 98μ.

1225μ(2 9.8μ)+125(2 9.8μ)^{2}=196μ

(2 9.8μ)[1225μ+125(2 9.8μ]=196μ

500 2450μ = 196μ

Hence, μ = 500/2646 = 250/1323.

If the lorry accelerated uniformly to 36 kmh^{-1} in 10 seconds, how far would the box slide?

Given that the lorry accelerates uniformly to a speed of `w` ms^{-1} in `t` seconds, before continuing at a constant speed, how far will the box slide in total?