Work Energy And Power MCQs for NEET — Physics Questions with Answers

$x={(t-3)}^2$ ⇒ $v=\frac{dx}{dt}=2(t-3)$

at $t=0$; $v_1=-6m/s$ and at $t=6\sec$, $v_2=6m\text{}/\text{}s$ 

so, change in kinetic energy $=W=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2=0$ 

While moving from (0,0) to (a,0)

Along positive x-axis, y = 0

$\therefore \overrightarrow{F}=-kx\hat{j}$

i.e. force is in negative y-direction while displacement is in positive x-direction.

∴ W₁ = 0

Because force is perpendicular to displacement

Then particle moves from (a, 0) to (a, a) along a line parallel to y-axis (x = +a) during this $\overrightarrow{F}=-k(y\hat{i}+a\hat{J})$

The first component of force, $-ky\hat{i}$ will not contribute any work because this component is along negative x-direction $(-\hat{i})$ while displacement is in positive y-direction (a,0) to (a,a). The second component of force i.e. $-ka\hat{j}$ will perform negative work

∴ $W_2=(-ka\hat{j})\left(a\hat{j}\right)$ = $(-ka)\left(a\right)\text{\hspace{0.17em}}=-k{a}^2$

So net work done on the particle W = W₁ + W₂

= $0+(-k{a}^2)=-k{a}^2$

Gain in potential energy $\Delta U=\frac{mgh}{1+\frac{h}{R}}$   

If h = R then $\Delta U=\frac{mgR}{1+\frac{R}{R}}=\frac{1}{2}mgR$

$\text{Stopping distance}=\frac{\text{kinetic energy}}{\text{retarding force}}$ ⇒ $s=\frac{1}{2}\frac{m{u}^2}{F}$

If lorry and car both possess same kinetic energy and retarding force is also equal then both come to rest in the same distance. 

Potential energy of the particle $U=k(1-e^{-x^2})$

Force on particle $F=\frac{-dU}{dx}=-k[-e^{-x^2}\times (-2x\left)\right]$

F $=\text{\hspace{0.17em}}-2kx{e}^{-x^2}=-2kx\left[1-x^2+\frac{{\displaystyle x^4}}{{\displaystyle 2!}}-.....\right]$

For small displacement $F=-2kx$

⇒ $F\left(x\right)\propto -x$ i.e. motion is simple harmonic motion. 

Kinetic energy acquired by the body

= Force applied on it × Distance covered by the body

K.E. = F × d

If F and d both are same then K.E. acquired by the body will be same

$P=Fv=mav=m\left(\frac{{\displaystyle dv}}{{\displaystyle dt}}\right)v$ ⇒ $\frac{P}{m}dt=vdv$

⇒ $\frac{P}{m}\times t=\frac{v^2}{2}$ ⇒ $v={\left(\frac{{\displaystyle 2P}}{{\displaystyle m}}\right)}^{1\text{}/\text{}2}{\left(t\right)}^{1/2}$

Now $s={\int }_{}^{}vdt={\int }_{}^{}{\left(\frac{2P}{m}\right)}^{1/2}{t}^{1/2}dt$

∴ $s={\left(\frac{{\displaystyle 2P}}{{\displaystyle m}}\right)}^{1/2}\left[\frac{{\displaystyle 2t^{3/2}}}{{\displaystyle 3}}\right]$ ⇒ $s\propto t^{3/2}$ 

Displacement of the particle,

      $$\begin{gathered} ∆\overrightarrow{s}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1 \\ =4\hat{j}+3\hat{k}-\left(-2\hat{i}+5\hat{j}\right) \\ =2\hat{i}-\hat{j}+3\hat{k} \end{gathered}$$

Force on the particle,

$$\begin{gathered} \overrightarrow{\mathrm{F}}=4\hat{\mathrm{i}}+3\hat{\mathrm{j}} \mathrm{N} \\ \therefore \mathrm{Work} \mathrm{done}, \mathrm{W}=\overrightarrow{\mathrm{F}}.∆\overrightarrow{\mathrm{s}} \\ =\left(4\hat{\mathrm{i}}+3\hat{\mathrm{j}}\right).\left(2\hat{\mathrm{i}}-\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right) \\ =8-3=5\mathrm{J} \end{gathered}$$

(c) According to question, a body of mass 1kg begins to move under the action of time dependent force,

    $F=\left(2t \hat{i}+3t^2 \hat{j}\right)$

where  $\hat{i}$ and $\hat{j}$ are unit vectors along X and Y axis

$$\begin{gathered} \because F=ma \\ \Rightarrow a=\frac{F}{m} \\ \Rightarrow a=\frac{\left(2t\hat{i}+3t^2\hat{j}\right)}{1} \left\{\because m=1 kg\right\} \\ \Rightarrow a=\left(2t\hat{i}+3t^2\hat{j}\right)m/s \end{gathered}$$

$\because acceleration, a =\frac{dv}{dt}$

$\Rightarrow dv=adt$                ...(i)

integrating both sides, we get

              $$\begin{gathered} \int dv=\int adt \\ =\int \left(2t\hat{i}+3t^2\hat{j}\right)dt \\ v=t^2\hat{i}+t^3\hat{j} \end{gathered}$$

$\because$Power developed by the force at the time t will be given by as

                            $$\begin{gathered} P=F.v=\left(2t\hat{i}+3t^2\hat{j}\right).\left(t^2\hat{i}+t^3\hat{j}\right) \\ =\left(2t.t^2+3t^2.t^3\right) \\ P=\left(2t^3+3t^5\right)W \end{gathered}$$

$$\begin{gathered} Given that; K_P>K_Q. \\ Case \left(a\right): \\ The elongation is same i.e. x_1=x_2=x \\ So, W_P=\frac{1}{2}{K}_P{x}^2 \& W_Q=\frac{1}{2}{K}_Q{x}^2 \\ So, W_P>W_Q\Rightarrow \frac{W_P}{W_Q}>1. \\ Case \left(b\right): \\ The spring force is same i.e. F_1=F_2=F. \\ So, x_1=\frac{F}{K_P} \& x_2=\frac{F}{K_Q} \\ \Rightarrow W_P=\frac{1}{2}{K}_P{x_1}^2=\frac{F^2}{2K_P} \& W_Q=\frac{1}{2}{K}_Q{x_2}^2=\frac{F^2}{2K_Q} \\ \Rightarrow W_P<W_Q\Rightarrow \frac{W_P}{W_Q}<1. \end{gathered}$$