You are an industrial engineer with a shipping company. As part of the package-handling system, a small box with mass \(1.60kg\)is placed against a light spring that is compressed\(0.280m\). The spring has force constant\(k = 45.0N/m\). The spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is\({\mu _k} = 0.300\). When the box has travelled\(0.280m\)and the spring has reached its equilibrium length, the box loses contact with the spring.

a. What is the speed of the box at the instant when it leaves the spring?

b. What is the maximum speed of the box during its motion?

The sum of an object's kinetic and potential energy changes represents the net work done by forces on the object.

\({K_i} + {U_i} + {W_{other}} = {K_f} + {U_f}\)

Where\(i\)stands for initial and\(f\)for final.

We know that,

\({K_i} + {U_i} + {W_{other}} = {K_f} + {U_f}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 1 \right)\)

Where\(i\)stands for initial and \(f\) for final.

Here in this case, other forces are just the friction.

Now, initially the spring is compressed by \(\Delta l = 0.28m\) and the box is at rest.

Hence \({K_i} = 0,\;\;\;\;\;\;\;\;\;\;{U_i} = \frac{1}{2}k{\left( {\Delta l} \right)^2}\)

The work done by friction from the moment the spring is releasewd to the moment it reaches equilibrium is

\({W_{other}} = {W_{friction}} = - {\mu _k}mg\Delta l\).

When the spring reaches equilibrium, \({U_f} = 0\).

Put all these values in equation (1)

\(\begin{aligned}{} \Rightarrow 0 + \frac{1}{2}k{\left( {\Delta l} \right)^2} - {\mu _k}mg\Delta l = {K_f} + 0\\ \Rightarrow {K_f} = \frac{1}{2}k{\left( {\Delta l} \right)^2} - {\mu _k}mg\Delta l\\ \Rightarrow \frac{1}{2}m{v_f}^2 = \frac{1}{2}k{\left( {\Delta l} \right)^2} - {\mu _k}mg\Delta l\\ \Rightarrow {v_f} = \sqrt {\frac{k}{m}{{\left( {\Delta l} \right)}^2} - 2{\mu _k}g\Delta l} \end{aligned}\)

\(\begin{aligned}{} \Rightarrow {v_f} = \sqrt {\frac{{45}}{{1.6}}{{\left( {0.28} \right)}^2} - 2\left( {0.3} \right)\left( {9.8} \right)\left( {0.28} \right)} \\ \Rightarrow {v_f} = 0.75\,{\rm{m/s}}\end{aligned}\)

Hence the speed of the box at the instant when it leaves the spring is \(0.75\,{\rm{m/s}}\).

Let \(x\) be the elongation of the spring.

We will find the relationship \(v\left( x \right)\), then find for which \(x\) this function attains maximum value and then insert this \(x\) to obtain the maximum speed of the box during this process.

If the elongation reduced from \( - \Delta l\;{\rm{to }}x\), the block traversed the distance of \(\Delta l - \left| x \right| = \Delta l + x\left( {x < 0} \right)\). Thus the work done by friction is \({W_{friction}} = - {\mu _k}mg\left( {\Delta l + x} \right)\).

Now, from equation (1),

\(\begin{aligned}{} \Rightarrow \frac{1}{2}k{\left( {\Delta l} \right)^2} - {\mu _k}mg\left( {\Delta l + x} \right) = \frac{1}{2}mv{\left( x \right)^2} + \frac{1}{2}k{x^2}\\ \Rightarrow v{\left( x \right)^2} = - \frac{k}{m}{x^2} - 2{\mu _k}g\left( {\Delta l + x} \right) + \frac{k}{m}{\left( {\Delta l} \right)^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 2 \right)\end{aligned}\)

Since the quadratic function is strictly increasing where \(v \ge 0\), function \(v\left( x \right)\) and \(v{\left( x \right)^2}\) attain their maximum value at the same point.

Now, \(f\left( x \right) = v{\left( x \right)^2}\) is a quadratic function in \(x\) and attains precisely one maximum at \({x_0}\).

The necessary condition for \(f\) to have a maximum at \({x_0}\) is,

\(\begin{aligned}{}f'\left( {{x_0}} \right) & = 0\\ \Rightarrow - \frac{{2k}}{m}{x_0} - 2{\mu _k}g & = 0\end{aligned}\)

\(\begin{aligned}{} \Rightarrow {x_0} & = \frac{{{\mu _k}gm}}{k}\\ \Rightarrow {x_0}& = - \frac{{\left( {0.3} \right)\left( {1.6} \right)\left( {9.8} \right)}}{{45}}\,m\\ \Rightarrow {x_0} & = - 10.5\,m\end{aligned}\)

Insert this in equation (2)

\(\begin{aligned}{}{v_{\max }} &= v\left( {{x_0}} \right)\\ \Rightarrow {v_{\max }}& = \sqrt { - \frac{{{\mu _k}^2{g^2}m}}{k} - 2{\mu _k}g\left( {\Delta l - \frac{{{\mu _k}gm}}{k}} \right) + \frac{k}{m}{{\left( {\Delta l} \right)}^2}\;} \\ \Rightarrow {v_{\max }} & = \sqrt {\frac{{{\mu _k}^2{g^2}m}}{k} - 2{\mu _k}g\left( {\Delta l} \right) + \frac{k}{m}{{\left( {\Delta l} \right)}^2}} \\ \Rightarrow {v_{\max }} & = \sqrt {\frac{{{{\left( {0.3} \right)}^2}{{\left( {9.8} \right)}^2}\left( {1.6} \right)}}{{45}} - 2\left( {0.3} \right)\left( {9.8} \right)\left( {0.28} \right) + \frac{{45}}{{1.6}}{{\left( {0.28} \right)}^2}} {\rm{m/s}}\\ \Rightarrow {v_{\max }} & = 0.93\,{\rm{m/s}}\end{aligned}\)

Hencethe maximum speed of the box during its motion is\(0.93\,{\rm{m/s}}\).