A 0.30 kgladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring $\left(k=500N/m\right)$ whose other end is fixed. The ladle has a kinetic energy of 10 J as it passes through its equilibrium position (the point at which the spring force is zero). (a) At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? (b) At what rate is the spring doing work on the ladle when the spring is compressed 0.10 mand the ladle is moving away from the equilibrium position?

The mass of the ladle m = 0.30 kg

The spring constant of the spring k = 500 N/m

The KE of the ladle at equilibrium (spring force = 0) is E = 10 J.

The total mechanical energy of the system is distributed as the kinetic energy of the ladle and the potential energy of the spring. The rate of doing work by the spring is the power due to the force of the spring and it can be calculated by using the product of force and speed of the ladle.

Formulae are as follows:

$$\begin{gathered} \mathrm{P}=\frac{\mathrm{W}}{\mathrm{t}} \\ =\mathrm{F}\mathrm{v} \end{gathered}$$ …(i)

KE of the ladle is given by

$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$ …(ii)

PE of spring is given by,

$\mathrm{PE}=\frac{1}{2}\mathrm{k}{\mathrm{x}}^2$ …(iii)

where, m is mass, v is velocity, F is force, w is work done, k is spring constant, x is displacement and tis time.

At the equilibrium position of the ladle, the spring force is zero. So, the work done by the spring is zero. Hence, the rate at which the spring is doing the work is also zero.

It is given that at the equilibrium position, the energy is 10 J, which is the KE of the ladle. Hence, this is the maximum energy the system of spring and the ladle can have.

Now, when the spring is compressed 0.10 m, this total energy will bethesum of KE of the ladle and the PE of the spring.

Hence,

$$\begin{gathered} \mathrm{E}=\mathrm{KE}+\mathrm{PE} \\ =\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\mathrm{k}{\mathrm{x}}^2 \\ =10\mathrm{J} \end{gathered}$$

Now, determine the speed of the ladle at this position so as to determine the rate of doing work. The speed can be determined as follows:

The PE of the spring is,

$$\begin{gathered} \frac{1}{2}\mathrm{k}{\mathrm{x}}^2=\frac{1}{2}\times 500\mathrm{N}/\mathrm{m}\times {\left(0.1\mathrm{m}\right)}^2 \\ =2.5\mathrm{J} \\ \frac{1}{2}\mathrm{m}{\mathrm{v}}^2+\frac{1}{2}\mathrm{k}{\mathrm{x}}^2=\mathrm{E} \\ \frac{1}{2}{\mathrm{mv}}^2=\mathrm{E}-\frac{1}{2}\mathrm{k}{\mathrm{x}}^2 \\ =10\mathrm{J}-2.5\mathrm{J} \\ =7.5J \end{gathered}$$

Solve the equation for v.

$$\begin{gathered} {\mathrm{v}}^2=\frac{7.5\mathrm{J}\times 2}{\mathrm{m}} \\ =\frac{15\mathrm{J}}{0.30\mathrm{kg}} \\ =50{\mathrm{m}}^2/{\mathrm{s}}^2 \\ \mathrm{v}=7.07\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence,$\mathrm{v}=7.07\mathrm{m}/\mathrm{s}$

And then the rate of doing work is,

$$\begin{gathered} \mathrm{P}=\mathrm{Fv} \\ =\left(\mathrm{kx}\right)\left(\mathrm{v}\right) \\ =\left(500\mathrm{N}/\mathrm{m}\times 0.10\mathrm{m}\right)\times \left(7.07\mathrm{m}/\mathrm{s}\right) \\ =353.5\mathrm{W} \\ =3.5\times {10}^2\mathrm{W} \end{gathered}$$

Therefore, at equilibrium position, there is no force by the spring. Hence, all the energy is expressed as the KE of the ladle. Later, when the spring is compressed, the same energy is distributed as the PE of the compressed spring and the KE of the ladle. This helps us determine the power of the spring when it is in a compressed state