An initially stationary 2.0kgobject accelerates horizontally and uniformly to a speed of10 m/sin3.0 s. (a) In that 3.0 sinterval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?

It is given that,

The mass of the object is m=2.0 kg.

The speed of the object is v= 10 m/s.

Time interval is t=3.0 s.

The problem deals with the power and the work done. Work, energy, and power are the fundamental concepts of physics. Work is the displacement of an object when force is applied on it. From the given values of mass mand speed v, find the work doneW.By finding the value of forceF, find theinstantaneous power Pand $\mathit{P}\mathbf{\text{'}}$due to the force at the end of the interval and at the end of the first half of the interval respectively.

Formulae are as follow:

The work done is

$W=\frac{1}{2}m{v}^2$

The instantaneous power is

$P=Fv$

The acceleration is

$a=\frac{v}{t}$

where, Fis force, d is displacement, P is the power, Wis the work done, vis velocity, ais an acceleration, t is the time and m is mass.

The work done is given by,

$$\begin{gathered} W=\frac{1}{2}m{v}^2 \\ W=\frac{1}{2}\times 2.0\times {10}^2 \\ W=1.0\times {10}^2\mathrm{J} \end{gathered}$$

Hence, the work doneWon the object by the force is $W=1.0\times {10}^2\mathrm{J}.$

The instantaneous power is given by,

$P=Fv$

But, according to Newton’s second law,

$$\begin{gathered} F=ma \\ =m\frac{v}{t} \end{gathered}$$

Substituting this,

$$\begin{gathered} P=m\frac{v}{t}v \\ P=m\frac{v^2}{t} \\ P=2.0\times \frac{{10}^2}{3.0} \\ P=67\mathrm{W} \end{gathered}$$

Hence, instantaneous power due to the force at the end of the interval is$P=67\mathrm{W}$

The velocity at$t\text{'}=1.5\mathrm{s}$is,

$v\text{'}=at\text{'}$

But,acceleration

$$\begin{gathered} a=\frac{v}{t} \\ =\frac{10}{3.0} \\ =3.3\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Substituting this,

$v\text{'}=3.3\times 1.5=5.0\mathrm{m}/\mathrm{s}$

Therefore, the instantaneous power is given by,

$P\text{'}=Fv\text{'}$

But, force

$$\begin{gathered} F=ma \\ =2.0\times 3.3 \\ =6.6\mathrm{N} \end{gathered}$$

Substituting this,

$$\begin{gathered} P\text{'}=6.6\times 5.0 \\ P\text{'}=33\mathrm{W} \end{gathered}$$

Hence, instantaneous power $P\text{'}$due to the force at the end of the first half of the interval is$P\text{'}=33\mathrm{W}$.