Work Power And Energy Practice Questions

Given, work done by both person is same, $\text{W}_{\text{A}}=\text{W}_{\text{B}}$ Direction of first force with displacement, $\theta _1$ = 45° Direction of first force with displacement, $\theta _2$ = 60° Ratio of force applied by person A to force applied by person B is $\text{F}_{\text{A}}: \text{F}_{\text{B}}=1: \\sqrt{x}$. Work done by both person is same, $\text{W}_{\text{A}}=\text{W}_{\text{B}}=$ $\Rightarrow \text{F}_{\text{A}} \text{d} \cos \theta _{1}=\text{F}_{\text{B}} \text{d} \cos \theta _{2}$ (\because distance is same for both = d) ${\text{F}_{\text{A}}}/{\text{F}_{\text{B}}}=\frac{\cos \theta _{1}}{\cos \theta _{2}}=\frac{\cos 60°}{\cos 45°}$ $\Rightarrow {\text{F}_{\text{A}}}/{\text{F}_{\text{B}}}={\frac{1}{2}}/{\frac{1}{\\sqrt{2}}}=\frac{1}{\\sqrt{2}}$ $\Rightarrow {\text{F}_{\text{A}}}/{\text{F}_{\text{B}}}=\frac{1}{\\sqrt{x}}=\frac{1}{\\sqrt{2}}$ $\Rightarrow x=2$ Thus, the value of x is 2.

Given, initial speed of block, $u=40 {\text{ms}}^{-1}$ Let total mass of block = m ∴ Broken masses, $m_{1}=m_{2}=\frac{m}{2}$ Final speed of $m_{1}, v_{1}=60 {\text{ms}}^{-1}$ As we know that, kinetic energy, ${\text{KE}}=1 / 2 {mv}^{2}$ ∴ Initial kinetic energy ${\text{KE}}_{i}=\frac{1}{2} m u^{2}$ $=\frac{1}{2} m(40)^{2}=800 m$ ...(i) By using law of conservation of momentum, $m u=m_{1} v_{1}+m_{2} v_{2}$ $\therefore v_{2}=\frac{m \times 40-m / 2 \times 60}{m / 2}$ $=\frac{10 m}{m / 2}=20 m s^{-1}$ So, final kinetic energy ${\text{KE}}_{\text{f}}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}$ $=\frac{1}{2} \frac{m}{2}(v_{1}^{2}+v_{2}^{2})=\frac{m}{4}(60^{2}+20^{2})$ $=m / 4(3600+400)$ $=1000 m$ ...(ii) ∴ Divide Eq (ii) by Eq. (i), we get ${{\text{KE}}_{\text{f}}}/{{\text{KE}}_{i}}={1000 {\text{m}}}/{800}=\frac{5}{4} {\text{m}}$ So, x = 5

${F}=(5 {y}+20) \hat{j}$ $\omega =\int F d y=\int _{0} ^{10}(5 y+20) d y$ $=(\frac{5 y^{2}}{2}+20 y)_{0}^{10}$ $=\frac{5}{2} \\times 100+20 \\times 10$ $=250+200=450 {J}$

Given, potential energy, $U=\;\frac{\alpha }{r^{10}}-\;\frac{\beta }{r^{5}}-3$ As we know for equilibrium, differentiation of potential energy with respect to distance $(\;\frac{d u}{d r})=0$ $\Rightarrow \;\; \;\frac{d}{d r}(\;\frac{\alpha }{r^{10}}-\;\frac{\beta }{r^{5}}-3)=0$ $\Rightarrow \;\;$ $\;\frac{d}{d r}(\alpha r^{-10}-\beta r^{-5}-3) \;\; =0$ $-10 \alpha r^{-11}+5 \beta r^{-6}-0 \;\; =0$ $\Rightarrow \;\; 10 \alpha r^{-11} \;\; =5 \beta r^{-6}$ $\Rightarrow \;\; 2 \alpha r^{-11} \;\; =\beta r^{-6}$ $\Rightarrow \;\; \;\frac{2 \alpha }{\beta }=\;\frac{r^{-6}}{r^{-11}}=r^{5} \Rightarrow r \;\; =(\;\frac{2 \alpha }{\beta })^{1 / 5}$ $\Rightarrow \;\;\;a=1$ $\;\Rightarrow \;\;\frac{d}{d r}(\;\frac{\alpha }{r^{10}}-\;\frac{\beta }{r^{5}}-3) \;\; =0$ $\Rightarrow \;\;\frac{d}{d r}(\alpha r^{-10}-\beta r^{-5}-3) \;\; =0$ $\Rightarrow \; \;\; -10 \alpha r^{-11}+5 \beta r^{-6}-0 \;\; =0$ $\Rightarrow \;10 \alpha r^{-11} \;\; =5 \beta r^{-6}$ $\Rightarrow \;2 \alpha r^{-11} \;\; =\beta r^{-6}$ $\Rightarrow \;\;\frac{2 \alpha }{\beta }=\;\frac{r^{-6}}{r^{-11}}=r^{5} \Rightarrow r \;\; =(\;\frac{2 \alpha }{\beta })^{1 / 5}$ $\;\Rightarrow \;a \;\; =1$

Let a block of mass M splits into two masses $m_1$ and $m_2$ in ratio 1: 2. Then, mass of smaller part, $m_{1}=\frac{\text{M}}{1+2}=\frac{\text{M}}{3}$ Mass of bigger part, $m_{2}=2 \text{M} / 3$ Given, speed of mass M, v = 40 m/s Speed of mass $m_{1}, v_{1}=60 {m} / {s}$ Let speed of mass $m_{2}, v_{2}=v$ Using conservation of linear momentum, ${p}_{i}={p}_{\text{f}}$ $\text{M} {v}=m_{1} {v}_{1}+m_{2} {v}_{2}$ $\text{M} \times 40=\frac{\text{M}}{3} \times 60+\frac{2 \text{M}}{3} \times v$ \Rightarrow = 30 m/s ∴ The fractional change in kinetic energy, $\frac{\Delta\text{K}}{\text{K}}={\text{K}_{\text{f}}-\text{K}_{i}}/{\text{K}_{i}}={\text{K}_{\text{f}}}/{\text{K}_{i}}-1$ $=\frac{\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}}{\frac{1}{2} \text{M} v^{2}}-1$ ={\frac{1}{2}\[\frac{\text{M}}{3} ×(60)^{2}+\frac{2 \text{M}}{3}×(30)^{2}]$}/{\frac{1}{2} \text{M} \times (40)^{2}}-1=\frac{9}{8}-1=\frac{1}{8}$

$m g \cos \theta =\frac{m v^{2}}{R}$ .......(1) $\cos \theta =\frac{{h}}{{R}}$ Energy conservation ${mg}\{{R}-{h}\}=\frac{1}{2} {mv}^{2}$ .....(2) from (1) & (2) $\Rightarrow{mg}\{\frac{{h}}{{R}}\}={2 {mg}\{{R}-{h}\}}/{{R}}$ ${h}=\frac{2 {R}}{3}=2 {m}$

$P=$ const. $P=F_{V}=\frac{m v^{2} d v}{d x}$ $\int _{0} ^{x} \frac{P}{m} d x=\int _{0} ^{v} v^{2} d v$ $\frac{P x}{m}=\frac{v^{3}}{3}$ $(\frac{3 P x}{m})^{1 / 3}=v=\frac{d x}{d t}$ $(\frac{3 P}{m})^{1 / 3} \int _{0} ^{t} d t=\int _{0} ^{x} x^{-1 / 3} d x$ $\Rightarrowx=$\frac{8 P}{9 m})^{1 / 2} t^{3 / 2}$

Given, mass of rolling ball, $m=0.5 {kg}$ Speed of ball, $u=20 {ms}^{-1}$ Before deflection, Initial kinetic energy, ${KE}_{1}=\;\frac{1}{2} m u^{2}$ $=\;\frac{1}{2} \times 0.5 \times (20)^{2}$ $=\;\frac{1}{2} \times 0.5 \times 400=\;\frac{1}{2} \times \;\frac{1}{2} \times 400=100 {J}$ It is given in the question that after deflection the ball moves with $5 \%$ of its initial kinetic energy ${KE}_{f}=5 \%$ of ${KE}_{ {i}} \Rightarrow {KE}_{f}=\;\frac{5}{100} \times 100=5 {J}$ If the final speed of the ball is $v {ms}^{-1}$, then ${KE}_{f}=\;\frac{1}{2} m v^{2}$ $\Rightarrow \;\; 5=\;\frac{1}{2} \times 0.5 \times v^{2} \Rightarrow 10=0.5 \times v^{2}$ $\Rightarrow \;\; v^{2}=\;\frac{10}{0.5}=\;\frac{100}{5} \Rightarrow v^{2}=20$ $\Rightarrow \;\; v=\sqrt{20}=4.47 {ms}^{-1}$ $\Rightarrow \;\; v=4.47 {ms}^{-1}$

We know that, Power $=$ force $\times$ velocity $[ \because$ Given, power $=$ constant $]$ $\therefore$ Force $\times$ velocity $=$ constant $\Rightarrow \;\; m a \times v=$ constant $\Rightarrow \;\; a \times v=$ constant $\Rightarrow \;\; (v \;\frac{d v}{d x})(\;\frac{d x}{d t})=C \Rightarrow ∫ v d v=∫ C d t$ $\Rightarrow \;\; \;\frac{v^{2}}{2}=C t \Rightarrow v=\sqrt{2 C t}$ $\Rightarrow \;\; \;\frac{d x}{d t}=\sqrt{2 C t}$ $[ \because v=\;\frac{d x}{d t}]$ $\Rightarrow \;\; ∫ d x=∫ \sqrt{2 Ctdt} \Rightarrow x=\sqrt{2 C}(\;{t^{3{/ } 2}}/{3 {/ }2})$ $\Rightarrow \;\; x \propto t^{3{/ }2}$

Given, The mass of the particle, $m=10 {kg}$ The speed of the particle at point $A, v_{A}=0 \frac{m}{s}$ The elevation of the point $A$ from the point $B, h_{A}=5+h_{B}$ Let's consider the speed of the particle at point $B=v_{B}$Using the law of conservation of energy. Energy at point $A=$ Energy at point $B$ $\;\frac{1}{2} m v_{A}^{2}+m g h_{A}=\;\frac{1}{2} m v_{B}^{2}+m g h_{B}$ Substituting the values in the above equations, we get $\;\frac{1}{2}(10)(0)^{2}+10 \times 10 \times (5+h_{B})=\;\frac{1}{2}(10) v_{B}^{2}+10 \times 10 \times h_{B}$ $v_{B}=10 {m} / {s}=x {m} / {s}$ $\therefore \;\; x=10$

Given, mass of the ball, $m=4 {kg}$ Velocity of the ball, $v=10 \frac{m}{s}$ Force constant, $k=100 \frac{N}{m}$ The length of the spring, $x=8 {m}$ Let $x$ is the compressed length of the spring. Using the work-energy theorem, "It states that kinetic energy of the ball is converted into the stored energy of spring". $\;\frac{m v^{2}}{2}=\;\frac{k x^{2}}{2}$ $\Rightarrow \;\; \;\frac{4(10)^{2}}{2}=\;\frac{(100) x^{2}}{2}$ $\Rightarrow x=2 m$ The final length (compressed) of the spring $=8-2=6 {m}$ Hence, the value of $x$ to the nearest integer is 6 .

Given, the mass of the body = m The height from which the body dropped = h The speed of the body when reached the ground, $\text{v}_{\text{f}}=0.8 \\sqrt{\text{gh}}$ Initial velocity of the body, v = 0 m / s Using the work-energy theorem, Work done by gravity + Work done by air-friction = Final kinetic energy - Initial kinetic energy $\text{W}_{{\text{mg} }}+\text{W}_{{\text{air}- \text{friction} }}=\frac{1}{2} \text{m} \text{v}_{\text{f}}^{2}-\frac{1}{2} \text{m} \text{v}_{\text{i}}^{2}$ Here, work done by gravity = mgh \Rightarrow $\text{m} \text{g} \text{h}+\text{W}_{{\text{air}-\text{friction} }}=\frac{1}{2} \text{m}(0.8 \\sqrt{\text{g} \text{h}})^{2}-\frac{1}{2} \text{m}(0)^{2}$ \Rightarrow $\text{W}_{{\text{air} - \text{fricion} }}=\frac{0.64 \text{m} \text{g} \text{h}}{2}-\text{m} \text{g} \text{h}$ = 0.32 mgh - mgh = -0.68 mgh The value of the work done by the air friction is -0.68 mgh.

Work done ($W$) by a variable force $\vec{F}$ is given by the line integral:
$W = \int_{r_i}^{r_f} \vec{F} \cdot d\vec{r}$

1. Set up the Integral

Given $\vec{F} = -x\hat{i} + y\hat{j}$ and the displacement vector $d\vec{r} = dx\hat{i} + dy\hat{j}$:
$\vec{F} \cdot d\vec{r} = (-x\hat{i} + y\hat{j}) \cdot (dx\hat{i} + dy\hat{j}) = -x\,dx + y\,dy$

2. Define the Limits

The particle moves from $A(1, 0)$ to $B(0, 1)$:

• $x$ varies from $1$ to $0$
• $y$ varies from $0$ to $1$

3. Calculate Work Done

$W = \int_{1}^{0} -x\,dx + \int_{0}^{1} y\,dy$

Evaluating the integrals:
$W = \left[ -\frac{x^2}{2} \right]_{1}^{0} + \left[ \frac{y^2}{2} \right]_{0}^{1}$

$W = \left( 0 - \left(-\frac{1}{2}\right) \right) + \left( \frac{1}{2} - 0 \right)$
Final Answer: 1

elevater moving with constant speed hence T = 6800 + 9200 + 6000T = 22000 Npower = T . v = 22000 × 3 = 66000 W

Mechanical energy conservation between A & P

${F}=200 {N} \;\;\;$ for $0 \le {x} \le 15$ $=200- \frac{100}{15}( {x}-15)\;\;\;$ for $15\le {x} < 30$ ${W}=∫{Fdx}$ $=∫_{0} ^{15} 200 {dx}+∫_{15} ^{30} (300-\frac{100}{15} {x} ) {dx}$ $=200 \\times 15+300 \\times 15-\ \frac{100}{15}$ $\times \frac{(30^{2}-15^{2} )}{2}$ $=3000+4500-2250$ $=5250 \ {J}$

$\frac{{d} {K}}{{dE}}= {P}= {\cos t}$ $\Rightarrow {K}= {Pt}= \frac{1}{2} \ {mV}^{2}$ $\therefore {V}=\sqrt{\frac{2 {Pt}}{ {m}}}=\frac{{ds}}{ {dt}}$ $\therefore {S}=\sqrt{\frac{2 {P}}{ {m}}} \frac{2}{3} {t}^{ \frac{3}{2}}$

${U}=-\;\frac{A}{r^{4}}+\;\frac{B}{r^{12}}$ ${F}=-\;\frac{d U}{d r}$ $\;\; =-(A(-6 r^{-7}))+B(-12 r^{-13})$ $\;\text{ for equilibrium, }\; F=0$ $\therefore 0=\;\frac{6 A}{r^{2}}-\;\frac{12 B}{r^{13}}$ $\Rightarrow \;\frac{6 A}{12 B}=\;\frac{1}{r^{6}}$ $\Rightarrow {r}=(\;\frac{2 B}{A})^{\;\frac{1}{6}}$ $\therefore {U}=-\;\frac{A}{\frac{2 B}{A}}+\;\frac{B}{(\;\frac{2 B}{A})^{2}}$ $\;\; =-\;\frac{A^{2}}{2 B}+\;\frac{A^{2}}{4 B}$ $\;\; =-\;\frac{A^{2}}{4 B}$