Question

The only force acting on a 2.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 4.0 m/sin the positive x direction and sometime later has a velocity of 6.0 m/sin the positive y direction. How much work is done on the canister by the 5.0 N force during this time?

Short answer

20 J of work is done on the canister by the 5.0 N force.

Step-by-step solution

Given data

Mass of the canister, m=2.0 kg.

Initial velocity is, $\overrightarrow{{\mathrm{v}}_{\mathrm{i}}}=4.0\hat{\mathrm{i}}\frac{\mathrm{m}}{\mathrm{s}}$.

The final velocity is, data-custom-editor="chemistry" $\overrightarrow{{\mathrm{v}}_{\mathrm{f}}}=6.0\hat{\mathrm{j}}\frac{\mathrm{m}}{\mathrm{s}}$.

Force is, F=5.0 N.

Understanding the concept

The problem deals with the work-energy theorem. It states that the net work done by the forces on an object equals the change in its kinetic energy. We have the initial velocity of the canister for the given time period. We also know the mass of the canister. From this, we can find the initial and final energy. Using the work-energy theorem, we can find the work done on the canister by 5.0 N force.

Formula:

$$\begin{gathered} \mathrm{K}.\mathrm{E}.=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{W}={\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{final}}-{\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{initial}} \end{gathered}$$

Calculate the initial and final kinetic energy of the canister

Initial kinetic energy can be calculated as,

${\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{initial}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2$

Substitute the values in the above expression, and we get,

$$\begin{gathered} {\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{initial}}=\frac{1}{2}\times 2\times 4^2 \\ {\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{initial}}=16\mathrm{J} \end{gathered}$$

Final kinetic energy can be calculated as,

${\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{final}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2$

Substitute the values in the above expression, and we get,

$$\begin{gathered} {\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{final}}=\frac{1}{2}\times 2\times 6^2 \\ {\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{final}}=36\mathrm{J} \end{gathered}$$

Calculate the work done on the canister

From the work-energy theorem, change in the kinetic energy will be equal to the work done, which can be written as,

$\mathrm{W}={\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{final}}-{\left(\mathrm{K}.\mathrm{E}.\right)}_{\mathrm{initial}}$

Substitute the values in the above expression, and we get,

data-custom-editor="chemistry" $\begin{array}{rcl}\mathrm{W}& =& 36\mathrm{J}-16\mathrm{J}\\ & =& 20\mathrm{J}\end{array}$

Thus, 20 J of work is done on the canister by the 5.0 N force.