In three situations, a briefly applied horizontal force changes the velocity of a hockey puck that slides over frictionless ice. The overhead views of $\mathbf{Fig}\mathbf{.}\mathbf{}\mathbf{7}\mathbf{-}\mathbf{17}$indicates for each situation, the puck’s initial speed $\mathit{v}$, its final speed $\mathit{v}$, and the directions of the corresponding velocity vectors. Rank the situations according to the work done on the puck by the applied force, most positive first and most negative last

1. An applied horizontal force changes the velocity of a hockey puck that slides over frictionless ice.
2. The initial speed of the puck is${\mathrm{v}}_1$.
3. The final speed of the puck is${\mathrm{v}}_{\mathrm{f}}$.
4. Graphs are given in Fig 7.17

We can rank the work done using the work-energy theorem, which states that work done is a change in kinetic energy. Using this concept, we can get the required values of the work done for all the individual cases of the changing velocities.

Formulae:

The kinetic energy of a body,

$\mathbf{K}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\times }\mathbf{m}\mathbf{\times }{\mathbf{v}}^{\mathbf{2}}$ (1)

The work done by a body,

$\mathbf{W}\mathbf{=}\mathbf{K}\mathbf{.}\mathbf{E}{\mathbf{.}}_{\mathbf{f}}\mathbf{-}\mathbf{K}\mathbf{.}{\mathbf{E}}_{\mathbf{.}\mathbf{i}}$ (2)

The work done is given using equation (1) in equation (2) as follows:

$\mathrm{W}=0.5\times \mathrm{m}\times {{\mathrm{v}}^2}_{\mathrm{f}}-0.5\times \mathrm{m}\times {{\mathrm{v}}^2}_{\mathrm{i}}..............\left(3\right)$

Assume the mass value as$\mathrm{m}=1\mathrm{kg}$

For graph A, the work done can be given by substituting the given values in equation (a) as follows:

role="math" localid="1657169976410" $$\begin{gathered} \mathrm{W}=0.5\times \left(1\mathrm{kg}\right)\times {(5\mathrm{m}/\mathrm{s})}^2-0.5\times \left(1\mathrm{kg}\right)\times {(6\mathrm{m}/\mathrm{s})}^2 \\ =-5.5\mathrm{J} \end{gathered}$$

For graph B, the work done can be given by substituting the given values in equation (a) as follows:

$$\begin{gathered} \mathrm{W}=0.5\times \left(1\mathrm{kg}\right)\times {(3\mathrm{m}/\mathrm{s})}^2-0.5\times \left(1\mathrm{kg}\right)\times {(4\mathrm{m}/\mathrm{s})}^2 \\ =-3.5\mathrm{J} \end{gathered}$$

For graph B, the work done can be given by substituting the given values in equation (a) as follows:

$$\begin{gathered} \mathrm{W}=0.5\times \left(1\mathrm{kg}\right)\times {(4\mathrm{m}/\mathrm{s})}^2-0.5\times \left(1\mathrm{kg}\right)\times {(2\mathrm{m}/\mathrm{s})}^2 \\ =6\mathrm{J} \end{gathered}$$

So, the rank of the positions according to the work done is $\mathrm{C}>\mathrm{B}>\mathrm{A}$.