91影视

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,

$\mathbf{K}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{脳}\mathbf{m}\mathbf{脳}{\mathbf{v}}^{\mathbf{2}}$ (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:

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

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

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

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

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

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

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