Question

Question: A 140Kghoop rolls along a horizontal floor so that the hoop’s center of mass has a speed of 0.150 m/s.How much work must be done on the hoop to stop it?

Short answer

Answer

Work needed to stop the hoop is -3.15j .

Step-by-step solution

Given 

$$\begin{gathered} m=140kg \\ v=0.150\text{m/s} \end{gathered}$$

To understand the concept

The work needed to stop the hoop is negative of the kinetic energy of the hoop.

As we get the kinetic energy, we can get the required answer.

The total kinetic energy of the hoop is given as-

$\mathbf{K}\mathbf{.}\mathbf{E}{\mathbf{.}}_{\mathbf{tot}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$

Calculate the work that must be done on the hoop to stop it

$$\begin{gathered} \text{K}.\text{E}.=\frac{1}{2}I{\omega }^2+\frac{1}{2}{\text{mv}}^2 \\ =\left(\frac{1}{2}m{r}^2\times \left(\frac{v^2}{r^2}\right)\right)+\frac{1}{2}m{v}^2 \\ =\frac{1}{2}\times 140\text{kg}\times {\left(0.150m/s\right)}^2+\frac{1}{2}\times 140\text{kg}\times {\left(0.150m/s\right)}^2 \\ =3.15\text{J} \end{gathered}$$

The final kinetic energy will be zero as it stops. So,

Work done will be-

$$\begin{gathered} W=\Delta K.E. \\ =K.E{.}_2-K.E{.}_1 \end{gathered}$$

For the given values, we have-

$$\begin{gathered} W=0-\left(3.15\text{J}\right) \\ =-3.15\text{J} \end{gathered}$$

Work done is negative as the force is applied in the opposite direction to the direction of motion. In other words, we will have to spend this energy to stop the hoop.