Question

A force of 5.0 Nacts on a 15.0 kgbody initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.

Short answer

1. The work done by the force in the first second of motion is 0.83 J
2. The work done by the force in the 2^nd second of motion is 2.5 J
3. The work done by the force in the third second of motion is 4.2 J
4. The instantaneous power at the end of third second is 5.0 W

Step-by-step solution

Given

1. The force acting on the body, F=15.0 N
2. The mass of the body, m=15 kg
3. The initial speed of the body, v₀=0 m/s

 Step 2: To understand the concept of work and power

The velocity at the end of each second can be determined using kinematical equations. The work done on a particle by a constant force during its displacement is given by,

$\mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{d}}$

The rate at whichtheforce does the work on the object is called as power due to the force. Power is given by,

${\mathbf{P}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{W}}{\mathbf{∆}\mathbf{t}}\mathbf{=}\mathbf{Fv}$

Formula:

$\begin{array}{rcl}∆\mathrm{x}& =& {\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2\\ \mathrm{W}& =& \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{x}}\\ {\mathrm{P}}_{\mathrm{avg}}& =& \frac{\mathrm{W}}{∆\mathrm{t}}\\ & =& \mathrm{Fv}\end{array}$

a) Calculate the work done in the 1stsecond

The force acting on the body will accelerate it. As a starting point, we will determine the speed of the body at the end ofthefirst second using the kinematical equation mentioned above.

The acceleration of the body will be

$$\begin{gathered} \mathrm{F}=\mathrm{ma} \\ \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

Substitute all the value in the above equation.

$\begin{array}{rcl}\mathrm{a}& =& \frac{\mathrm{F}}{\mathrm{m}}\\ & =& \frac{5.0\mathrm{N}}{15\mathrm{kg}}\\ & =& 0.33\mathrm{m}/{\mathrm{s}}^2\end{array}$

Now, the distance covered in the first second of motion will be,

$\begin{array}{rcl}∆{\mathrm{x}}_1& =& {\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2\\ & =& 0+\frac{1}{2}\times 3.33\mathrm{m}/{\mathrm{s}}^2\times 1\mathrm{s}\\ & =& 0.165\mathrm{m}\end{array}$

Hence, the work done in the first second will be,

$\begin{array}{rcl}\mathrm{W}& =& \overrightarrow{\mathrm{F}}\overrightarrow{∆\mathrm{d}}\\ & =& \mathrm{F}∆\mathrm{x}\\ & =& 5.0\mathrm{N}\times 0.165\mathrm{m}\\ & =& 0.83\mathrm{J}\end{array}$

Therefore, work done in the first second is 0.83 J

b) Calculate the work done in the 2nd second

Now, we will determine the speed of the body at the end ofthe2^nd second using the kinematical equation mentioned above.

The acceleration of the body 0.33 m/s²

Now, the distance covered in two seconds of motion will be,

$\begin{array}{rcl}∆{\mathrm{x}}_2& =& {\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2\\ & =& 0+\frac{1}{2}\times 0.33\mathrm{m}/{\mathrm{s}}^2\times {\left(2\mathrm{s}\right)}^2\\ & =& 0.66\mathrm{m}\end{array}$

The distance covered in the 2^nd second of the motion will be,

$\begin{array}{rcl}∆\mathrm{x}& =& ∆{\mathrm{x}}_2-∆{\mathrm{x}}_1\\ & =& 0.66\mathrm{m}-0.165\mathrm{m}\\ & =& 0.495\mathrm{m}\end{array}$

Hence, the work done in the 2^nd second will be,

$\begin{array}{rcl}\mathrm{W}& =& \overrightarrow{\mathrm{F}}\overrightarrow{∆\mathrm{d}}\\ & =& \mathrm{F}∆\mathrm{x}\\ & =& 5.0\mathrm{N}\times 0.495\mathrm{m}\\ & =& 2.47\mathrm{J}\\ & =& 2.5\mathrm{J}\end{array}$

Therefore, work done in the 2^nd second is 2.5 J.

c) Calculate the work done in the 3rd second

We will now determine the speed of the body at the end of the third second using the kinematical equation mentioned above.

The acceleration of the body =0.33 m/s²

Now, the distance covered in three seconds of motion will be

$\begin{array}{rcl}∆{\mathrm{x}}_3& =& {\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2\\ & =& 0+\frac{1}{2}\times 0.33\mathrm{m}/{\mathrm{s}}^2\times {\left(3\mathrm{s}\right)}^2\\ & =& 1.49\mathrm{m}\end{array}$

The distance covered in the third second of the motion will be,

$\begin{array}{rcl}∆\mathrm{x}& =& ∆{\mathrm{x}}_3-∆{\mathrm{x}}_2\\ & =& 1.49\mathrm{m}-0.66\mathrm{m}\\ & =& 0.83\mathrm{m}\end{array}$

Hence, the work done in the third second will be,

$\begin{array}{rcl}\mathrm{W}& =& \overrightarrow{\mathrm{F}}\overrightarrow{∆\mathrm{d}}\\ & =& \mathrm{F}∆\mathrm{x}\\ & =& 5.0\mathrm{N}\times 0.83\mathrm{m}\\ & =& 4.2\mathrm{J}\end{array}$

Therefore, work done in the 3^rd second is 4.2 J.

d) Calculate the instantaneous power due to the force at the end of the third second

The velocity at the end of third second will be,

$\begin{array}{rcl}\mathrm{v}& =& {\mathrm{v}}_0+\mathrm{at}\\ & =& 0+0.33\mathrm{m}/{\mathrm{s}}^2\times 3\mathrm{s}\\ & =& 0.99\mathrm{m}/\mathrm{s}\\ & =& 1.0\mathrm{m}/\mathrm{s}\end{array}$

The instantaneous power will be,

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{avg}}& =& \frac{\mathrm{W}}{∆\mathrm{t}}\\ & =& \mathrm{Fv}\\ & =& 1.0\mathrm{N}\times 1.0\mathrm{m}/\mathrm{s}\\ & =& 5.0\mathrm{W}\end{array}$

Therefore, the instantaneous power at the end of the third second is 5.0 W.