Work Power And Energy Practice Questions

$\;\;\text{ Given }\; {\text{m}}=100 {\text{kg}}$ $\; {\ s}=10 {\text{m}}$ $\;\mu =0.4$ $\;\;\text{ As }\; {\text{f}}=\mu {\text{mg}}=0.4 \times 100 \times 10=400 {\text{N}}$ $\;\;\text{ Now }\; {\text{W}}= {\text{f}} . {\ s}=400 \times 10=4000 {\text{J}}$

Work done by gravity is independent of path. It depends only on vertical displacement so work done in both cases will be same.Option (1) is correct

$\;\;\text{ Kinetic energy }\;(K)=\;\frac{P^2}{2 m}$ $\;\Rightarrow {\text{P}}=\sqrt{2 {\text{mK}}}$ $\;\;\text{ If }\; {\text{K}}_{ {\text{f}}}=36 {\text{K}}_{ {\text{i}}}$ $\;\;\text{ So, }\; {\text{P}}_{ {\text{f}}}=6 {\text{P}}_{ {\text{i}}}$ $\;\% \;\text{ increase \in momentum }\;=\;{ {\text{P}}_{ {\text{f}}}- {\text{P}}_{ {\text{i}}}}/{ {\text{P}}_{ {\text{i}}}} \times 100 \%$ $\;=\;{6 {\text{P}}_{ {\text{i}}}- {\text{P}}_{ {\text{i}}}}/{ {\text{P}}_{ {\text{i}}}} \times 100 \%$ $\;=500 \%$ $\;$

Given $U=2 x^{2+}3 y^{3+}2 z$ ${\text{F}}_{ {\text{x}}}=-\;{∂ {\text{U}}}/{∂ {\text{x}}}=-4 {\text{x}}$At $x=1$ magnitude of $F_x$ is $4 N$

$\; {\text{v}}=\alpha \sqrt{ {\text{x}}}$ $\;\;\text{ at }\; {\text{x}}=0: {\text{v}}=0$ $\;\ \;\text{ at }\; {\text{x}}= {\text{d}} ; {\text{v}}=\alpha \sqrt{ {\text{d}}}$ $\;\;\text{ W.D }\;= {\text{K}}_{ {\text{f}}}- {\text{K}}_{ {\text{i}}}$ $\;\;\text{ W.D }\;=\;\frac{1}{2} {\text{m}}(\alpha \sqrt{ {\text{d}}})^{2-}\;\frac{1}{2} {\text{m}}(0)^2$ $\;\Rightarrow \;\text{ W.D }\;=\;{ {\text{m}} \alpha ^2 {\text{d}}}/{2}$

$\;v^{2-}u^2=2 a S$ $\; (\;\frac{2 u}{3}) ^2=u^{2+}2(-a) (4 \times 10^{-2})$ $\;\;\frac{4 u^2}{9}=u^{2-}2 a (4 \times 10^{-2})$ $\;-\;\frac{5 u^2}{9}=-2 a (4 \times 10^{-2}) . . .(1)$ $\;0= (\;\frac{2 u}{3}) ^{2+}2(-a)(x)$ $\;-\;\frac{4 u^2}{9}=-2 a x . . .(2)$(1) /(2)$\;\frac{5}{4}=\;\frac{4 \times 10^{-2}}{x}$ $\;x=\;\frac{16}{5} \times 10^{-2}$ $\;x=3 \cdot 2 \times 10^{-2} {\text{m}}$ $\;x=32 \times 10^{-3} {\text{m}}$Note : Since no option is matching, Question should be bonus.

Velocity just before collision $=\sqrt{2 {\text{gh}}}$Velocity just after collision $=\sqrt{2 g (\;\frac{h}{2}) }$ $\therefore \Delta {\text{KE}}=\;\frac{1}{2} {\text{m}}(2 {\text{gh}})-\;\frac{1}{2} {\text{mgh}}$ $=\;\frac{1}{2} {\text{mgh}}$ $\therefore \%$ loss in energy$=\;{\Delta {\text{KE}}}/{ {\text{KE}}_{ {\text{i}}}} \times 100=\;{\;\frac{1}{2} {\text{mgh}}}/{\frac{1}{2} {\text{mg}} 2 {\text{h}}} \times 100=50 \%$Hence option (4)

\;{ {\text{F}}}^{\to }= (6 {\text{t}} { {\text{i}}^{\^}}+6 {\text{t}}^2 { {\text{j}}^{\^}}) {\text{N}} \;{ {\text{F}}}^{\to }= {\text{ma}}= (6 {\text{t}} { {\text{i}}^{\^}}+6 {\text{t}}^2 { {\text{j}}^{\^}}) \;{ {\text{a}}}^{\to }=\;{{ {\text{F}}}^{\to }}/{ {\text{m}}}= (3 {\text{t}} { {\text{i}}^{\^}}+3 {\text{t}}^2 { {\text{j}}^{\^}}) \;{ {\text{v}}}^{\to }=∫_{0}^{}{ {\text{t}}} { {\text{a}}}^{\to } {\text{dt}}=\;{3 {\text{t}}^2}/{2} { {\text{i}}^{\^}}+ {\text{t}}^3 { {\text{j}}^{\^}} $\; {\text{P}}={ {\text{F}}}^{\to } \cdot { {\text{v}}}^{\to }= (9 {\text{t}}^{3+}6 {\text{t}}^5) {\text{W}}$

Apply W.E.T. from A to B$\;\Rightarrow {\text{W}}_{ {\text{mg}}}= {\text{K}}_{ {\text{B}}}- {\text{K}}_{ {\text{A}}}$ $\;\Rightarrow {\text{mg}} \times (\;{ {\text{R}}}/{{\sqrt{2}}}+ {\text{R}}) =\;\frac{1}{2} {\ mv}_{ {\text{B}}}^{2-}0 \{ {\text{v}}_{ {\text{A}}}=0 \;\text{ rest }\;\}$ $\;\Rightarrow {\text{mgR}} \;{({\sqrt{2}}+1)}/{{\sqrt{2}}}=\;\frac{1}{2} {\ mv}_{ {\text{B}}}^2$ $\;\Rightarrow \sqrt{ {\text{gR}} \;{2({\sqrt{2}}+1)}/{{\sqrt{2}}}}= {\text{v}}_{ {\text{B}}}$ $\;\Rightarrow \sqrt{\;\frac{10 \times 14 \times 2(2.4)}{1.4}}= {\text{v}}_{ {\text{B}}}$ $\;\Rightarrow 21.9= {\text{v}}_{ {\text{B}}}$Hence option (2) is correct

$\; {\text{KE}}=\;{ {\text{P}}^2}/{2 {\text{m}}}$ $\; {\text{P}} \propto \sqrt{ {\text{m}}}$Hence, $P_A: P_B: P_C$ $={\sqrt{400}}: {\sqrt{1200}}: {\sqrt{1600}}=1: {\sqrt{3}}: 2$

By COME$\; {\text{KE}}_{ {\text{A}}}+ {\text{U}}_{ {\text{A}}}= {\text{KE}}_{ {\text{B}}}+ {\text{U}}_{ {\text{B}}}$ $\;0+ {\text{mg}}(1)=\;\frac{1}{2} {\ mv}^{2+} {\text{mg}} \times 0.5$ $\;v={\sqrt{g}}={\sqrt{10}} {\text{m}} / {\ s}$

Work done again frictional force$\;=\mu {\text{N}} \times 10$ $\;=0.1 \times 5 \times 10=5 {\text{J}}$

${\text{KE}}=\;{ {\text{p}}^2}/{2 {\text{m}}}$Same momentum, so less mass means more KE.So $\;{ {\text{m}}}/{2}$ will have max. ${\text{KE}}$.

$\; {\text{w}}_{ {\text{g}}}+ {\text{w}}_{ {\text{Fr}}}+ {\text{w}}_{ {\ s}}=\Delta {\text{KE}}$ $\;5 \times 10 \times 5-0.5 \times 5 \times 10 \times {\text{x}}-\;\frac{1}{2} {\text{Kx}}^2=0-0$ $\;250=25 {\text{x}}+50 {\text{x}}^2$ $\;2 {\text{x}}^{2+} {\text{x}}-10=0$ $\; {\text{x}}=2$

Loss in kinetic energy $=$ Gain in potential energy $\;\Rightarrow \;\frac{1}{2} {\ mv}^2= {\text{mg}} ℓ(1-\cos \theta )$ $\;\Rightarrow \;{ {\text{v}}^2}/{ℓ}=2 {\text{g}}(1-\cos \theta )$Acceleration at lowest point $=-\;{ {\text{v}}^2}/{ℓ}$Acceleration at extreme point $=g sin \;\theta$ $\;\;\text{ Hence, }\; \;{ {\text{v}}^2}/{ℓ}= {\text{g}} sin \;\theta$ $\; \therefore sin \;\theta =2(1-\cos \theta )$ $\;\Rightarrow \tan \;\frac{\theta }{2}=\;\frac{1}{2} \Rightarrow \theta =2 \tan ^{-1} (\;\frac{1}{2})$

$\; {\text{P}}=\;\text{ constant }\; \Rightarrow {\text{FV}}=\;\text{ constant }\;$ $\;\Rightarrow {\text{m}} \;{ {\text{dV}}}/{ {\text{dt}}} {\text{V}}=\;\text{ constant }\;$ $\;∫_{0}^{ {\text{V}}} {\text{VdV}}=( {\text{C}}) ∫_{0}^{}{ {\text{t}}} {\text{dt}}$ $\; (\;{ {\text{V}}^2}/{2}) = {\text{Ct}}$ $V \propto t^{1 / 2}$ $\; {\text{S}}= {\text{K}} \times \;\frac{2}{3} {\text{t}}^{3 / 2}$ $\; {\text{S}} \propto {\text{t}}^{3 / 2}$ $\; \therefore \;\text{ displacement is proportional to }\;( {\text{t}})^{3 / 2}$

$\;{ {\text{F}}}^{\to }=(2+3 {\text{x}}) {\text{i}}$ $\; {\text{w}}=∫_{0}^{4} {\text{F}} \cdot {\text{dx}}=∫_{0}^{4}(2+3 {\text{x}}) \cdot {\text{dx}}$ $\; {\text{w}}= (2 {\text{x}}+\;{3 {\text{x}}^2}/{2}) ^4=(8+24)$ $\; {\text{w}}=32 {\text{J}}$

$\;\;\text{ Given, }\; a=-2 x$ $\;\Rightarrow \;\frac{v d v}{d x}=-2 x$ $\;\Rightarrow v d v=-2 x \text{dx}$ $\;\Rightarrow ∫_{v_1}^{}{v_2} v d v=-2 ∫_{0}^{x} x \text{dx}$ $\;\Rightarrow \;\frac{v_2^2}{2}-\;\frac{v_1^2}{2}=-\;\frac{2 x^2}{2}$ $\;\Rightarrow \;\frac{m v_1^2}{2}-\;\frac{m v_2^2}{2}=m x^2=\;\frac{10}{1000} x^2=10^{-2} x^2= (\;\frac{10}{x}) ^{-2}$ $\; {\text{n}}=2 .$