Work, Energy and Power

$$\begin{gathered} \mathrm{Using} \mathrm{the} \mathrm{formula} \mathrm{for} \mathrm{work} \mathrm{done} \\ \mathrm{W}=\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{r}} \\ =\overrightarrow{\mathrm{F}}.\left(\overrightarrow{{\mathrm{r}}_2}-\overrightarrow{{\mathrm{r}}_1}\right)=\left(4\hat{\mathrm{i}}+\hat{\mathrm{j}}+3\hat{\mathrm{k}} \right). \left\{\left(14\hat{\mathrm{i}}+13\hat{\mathrm{j}}+9\hat{\mathrm{k}}\right)-\left(3\hat{\mathrm{i}}+2\hat{\mathrm{j}}-6\hat{\mathrm{k}}\right)\right\} \\ =\left(4\hat{\mathrm{i}}+\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right). \left[14\hat{\mathrm{i}}+13\hat{\mathrm{j}}+9\hat{\mathrm{k}}-3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+6\hat{\mathrm{k}}\right] \\ =\left(4\hat{\mathrm{i}}+\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right). \left(11\hat{\mathrm{i}}+11\hat{\mathrm{j}}+15\hat{\mathrm{k}}\right) \\ \Rightarrow \mathrm{W} =44+11+45 \\ \mathrm{We} \mathrm{know} \mathrm{that} \left(\mathrm{i}.\mathrm{i}=\mathrm{j}.\mathrm{j}=\mathrm{k}.\mathrm{k}=1\right) \\ \Rightarrow \mathrm{W}=100\mathrm{J} \end{gathered}$$

Here :- Weight of block;ω = 2kN; Distance d =10m

Angle of inclination on the plane α =15⁰

The block will be pulled up on a smooth plane

Hence,force of resistance due to the inclination 

F =ωsinα = 2×10³×Sin 15⁰

=2×10³×0.2588

=0.5176 kN

Now work done W = force×displacement =0.1576×10³×10=5.17kJ

The spring force is an example of variable force which is conservative.

$$\begin{gathered} \mathrm{The} \mathrm{energy} \mathrm{stored} \mathrm{in} \mathrm{the} \mathrm{spring} \mathrm{is} \\ \mathrm{E}=\frac{1}{2}{\mathrm{kx}}^2 \\ \mathrm{and} \mathrm{expansion} \mathrm{in} \mathrm{the} \mathrm{spring} \mathrm{is} \\ \mathrm{x}=\frac{\mathrm{T}}{\mathrm{k}} \\ \mathrm{From} \mathrm{equations} \left(1\right) \mathrm{and}\left(2\right), \mathrm{we} \mathrm{get} \\ \mathrm{E}=\frac{1}{2}\mathrm{k}\frac{{\mathrm{T}}^2}{{\mathrm{K}}^2}=\frac{1}{2}\frac{{\mathrm{T}}^2}{\mathrm{k}}=\frac{{\mathrm{T}}^2}{2\mathrm{k}} \end{gathered}$$

From conservation of energy

Potential energy = translational kinetic energy + rotational kinetic energy

 mgh = $\frac{1}{2} {\mathrm{mv}}^2 + \frac{1}{2}$Iω²

  mgh  =  $\frac{1}{2} {\mathrm{mv}}^2 + \frac{1}{2} \left(\frac{2}{5}\right) {\mathrm{mR}}^2 \frac{{\mathrm{v}}^2}{{\mathrm{R}}^2}$

⇒ $\frac{7}{10} {\mathrm{mv}}^2$ = mgh

     v² =$\frac{10}{7}$gh

⇒    v = $\sqrt{\frac{10}{7}}$ gh

Hence too climb the inclined surface v should be  ≥ $\sqrt{\frac{10}{7}\mathrm{gh}}$

Heat added to the system ΔQ =110J

Internal energy Δu =40J

External work done is given by

 ΔW= ΔQ-Δu=110-40=70J

The velocity of second piece can be find out using conservation of momentum.

        0 + m₁ v₁ + m₂ v₂ = 2 × v₁ + 1 × 80

⇒         v₁ = - 40 m/s

Negative sign showing that particle is moving in opposite direction of othr particle direction.

The energy imparted to the fragments are converted into their kinetic energy.

∴   Total energy = $\frac{1}{2} {\mathrm{m}}_1 {\mathrm{v}}_1^{ 2} + \frac{1}{2} {\mathrm{m}}_2 {\mathrm{v}}_2^{ 2}$

                         = $\frac{1}{2} \times 2 \times 1600 + \frac{1}{2} \times 1 \times 6400$

                          = 4800 J

     Total energy = 4.8 kJ

When a cone filled with water is revolved in a vertical circle, then the velocity at the highest point is given by

$$\begin{gathered} \mathrm{v} \ge \sqrt{\mathrm{gr}} \\ \mathrm{v} = \mathrm{r\omega } = \mathrm{r}\frac{2\mathrm{\pi }}{\mathrm{T}} \end{gathered}$$

Hence, $\frac{2\mathrm{r\pi }}{\mathrm{T}} \ge \sqrt{\mathrm{rg}}$

$\mathrm{T} \le \frac{2\mathrm{\pi r}}{\sqrt{\mathrm{rg}}} = 2\mathrm{\pi }\sqrt{\frac{\mathrm{r}}{\mathrm{g}}}$

$$\begin{gathered} \le 2 \times 3.14 \times 0.638 \\ \le 4.0 \sec \\ \therefore {\mathrm{T}}_{\max } = 4 \sec \end{gathered}$$

According to law of conservation of momentum

          Iω = consant

I = moment of inertia

ω = angular speed

When viscous fluid of mass m is dropped and start spreading out then its moment of inertia increases and angular velocity decreases. But when it start falling then its moment of inertia again starts decreasing and angular velocity increases.

Amount of work done is given by

W=200cal= 200×4.2 J= 840 J       ($\because$1 cal=4.2J)

According to work-energy theorem,

  W = Δ K.E

⇒   W = $\frac{1}{2} \mathrm{m} {\mathrm{v}}_2^{ 2} - \frac{1}{2} \mathrm{m} {\mathrm{v}}_1^{ 2}$

⇒   F · dx  = K.E_f $- \frac{1}{2} \times 10 \times 100$

⇒   F · dx = K.E_f  $-$ 500

⇒   ${\int }_{30}^{20} - 0.1 \mathrm{x} \mathrm{dx}$ = K.E_f $-$ 500

⇒  $-0.1 {\left[\frac{{\mathrm{x}}^2}{2}\right]}_{20}^{30}$ = K.E_f $-$ 500

⇒   $-0.1 \left[\frac{900 - 400}{2}\right]$ = K.E_f $-$ 500

⇒   K.E_f = 500 $-$ 25

⇒    K.E_f = 475 J