Work Power And Energy Practice Questions

$\;F=5+3 y^2$ $\;W=∫_{2}^{5} (5+3 y^2) d y$ \;= \[5 y+;\frac{3 y^3}{3}]$ _2^5$$;=132 {\text{J}}$

$\;( {\text{KE}})=\;{ {\text{P}}^2}/{2 {\text{M}}}$ $\;\Rightarrow \;\frac{1}{2} {\mu}^2=\;\frac{(10)^2}{2 \times 5}$ $\;=\;\frac{1}{2} \times 5 \times {\text{u}}^2=\;\frac{100}{10}$ $\;\;\text{ Initial speed }\; {\text{u}}=2 {\text{m}} / {\ s}$ $\;\Delta {\text{KE}}= {\text{w}}_{\;\text{all forces }\;}$ $\;={ {\text{F}}}^{\to } \cdot { {\text{S}}}^{\to } \;\; (\theta =0^{\circ})$ $\;= {\text{F}} ( {\text{ut}}+\;\frac{1}{2} {\text{at}}^2)$ $\;=2 \cdot [2 \times 5+\;\frac{1}{2} \times \;\frac{2}{5} \times 5^2]$ $\;=30 {\text{J}}$

$\;\;\frac{1}{2} {\text{mV}}^2= {\text{Pt}}$ $\; {\text{V}}=\sqrt{\;{2 {\text{Pt}}}/{ {\text{m}}}}$ $\;\;{ {\text{dx}}}/{ {\text{dt}}}=\sqrt{\;{2 {\text{Pt}}}/{ {\text{m}}}}$ \; {\text{x}}=\sqrt{\;{2 {\text{P}}}/{ {\text{m}}} \;\frac{2}{3}} \[ {\text{t}}^{3 / 2}]$ _0^4$$; {\text{x}}=;{16 \sqrt{ {\text{P}}}}/{3}=;\frac{1}{3} \times 16 \sqrt{ {\text{P}}}$$;\alpha =4$

Let $u$ and $v$ be speeds, just before and after body strikes the ground.Given $\;{ {\text{u}}}/{ {\text{v}}}=\;\frac{4}{1}$ $\;\;\text{ loss \in }\; {\text{KE}}: \Delta {\text{KE}}=\;\frac{\;\frac{1}{2} {\mu}^2=\frac{1}{2} {\ mv}^2}{\frac{1}{2} {\mu}^2}$ $\;\Delta {\text{KE}}=1- (\;{ {\text{v}}}/{ {\text{u}}}) ^2=1-\;\frac{1}{16}=\;\frac{15}{16}$ $\;\;\text{ Percentage loss }\;=\;\frac{15}{16} \times 100=375$

$\; {\text{R}}_{\;\text{arg }\;}=15 {\text{cm}}=.15 {\text{m}}$ $\;\;\text{ By WET }\;$ $\; {\text{W}}_{ {\text{f}}}+ {\text{W}}_{\;\text{gravity }\;}=\Delta {\text{K}}= {\text{Kf}}- {\text{Ki}}$ $\; {\text{W}}_{ {\text{f}}}+10 \times .3=0-\;\frac{1}{2} \times 1 \times (22)^2$ $\; {\text{W}}_{ {\text{f}}}=-3-\;\frac{484}{2}=3-242=-245$Work by friction $=-245$ By NTA $(+245)$

$\; {\text{F}}=0.04 \times 500 \times 9.8$ $\;=20 \times 9.8=196$ $\; {\text{WD}}= {\text{F}} \times {\text{disc}}$ $\;=196 \times 4000$ $\;=784 {\text{kJ}}$

Statement-1 trueStament-2 false

Apply momentum conservation just before and just after the collision$\;0.1 \times 400=(3.9+.1) {\text{u}}$ $\;\Rightarrow {\text{u}}=10 {\text{m}} / {\ s}$ $\;\Delta {\text{KE}}= {\text{w}}_{\;\text{all FORCE }\;}$ $\; \because {\text{f}}=\mu {\text{mg}} \;\text{ (kinetic friction) }\;$ $\;\Rightarrow 0-\;\frac{1}{2}(4)(10)^2=-\mu (4) {\text{g}} \times 20$ $\;\Rightarrow \mu =0.25$

${\text{v}}^2\;= {\text{u}}^{2+}2 {\text{as}}$ $\;=2^{2+}2(2)(6)$ $\;=4+24=28$ ${\text{KE}}\;=\;\frac{1}{2} {\ mv}^2$ $\;=\;\frac{1}{2}(500) 28$ $\;=7000 {\text{J}}$ $\;=7 {\text{kJ}}$Ans. 7

$\;\;\frac{1}{2} \times 2 \times v^2=10000$ $\;\Rightarrow {\text{v}}^2=10000$ $\;\Rightarrow {\text{v}}=100 {\text{m}} / {\ s}$ $\;\Rightarrow {\text{v}}= {\text{at}}= {\text{a}} \times 5=100$ $\;\Rightarrow {\text{a}}=20 {\text{m}} / {\ s}^2$ $\; {\text{F}}= {\text{ma}}=2 \times 20=40 {\text{N}}$ $\;$

{ {\text{F}}}^{\to }= {\text{ti}}+3 {\text{t}}^2 \;{ {\text{j}}}^{\^} \;{ {\text{md}} { {\text{v}}}^{\to }}/{ {\text{dt}}}= {\text{t}} \;{ {\text{i}}}^{\^}+3 {\text{t}}^2 \;{ {\text{j}}}^{\^} {\text{m}}=1 {\text{kg}}, ∫_{0}^{{ {\text{v}}}^{\to }} {\text{dv}}=∫_{0}^{ {\text{t}}} {\text{tdt}} \;{˙{i}}^{\^}+∫_{0}^{ {\text{t}}} 3 {\text{t}}^2 {dt \;{j }^{\^} { {\text{v}}}^{\to }=\;{ {\text{t}}^2}/{2} \;{ {\text{i}}}^{\^}+ {\text{t}}^3 \;{ {\text{j}}}^{\^} $\;\text{ Power }\;={ {\text{F}}}^{\to } \cdot { {\text{V}}}^{\to }=\;{ {\text{t}}^3}/{2}+3 {\text{t}}^5$ $\;\text{ At }\; {\text{t}}=2, \;\text{ power }\;=\;\frac{8}{2}+3 \times 32$ $=100$

$\;W=\vec{F} \cdot (\vec{r}_f-\vec{r}_1)$ $\;=(5 {i}^{\wedge }+2 {j}^{\wedge }+7 {k}^{\wedge }) \cdot ((5 {i}^{\wedge }-2 {j}^{\wedge }+{k}^{\wedge })-(2 {i}^{\wedge }+3 {j}^{\wedge }-4{k}^{\wedge }))$ $\;W=40 {\text{J}}$

$\; {\text{P}}=\;\frac{\;\text{ Work }\;}{\;\text{ Time }\;}$ $\; {\text{P}}_1=\;{ {\text{mgh}}}/{ {\text{t}}_1}=\;{(300) {\text{g}}(100)}/{5}$ $\; {\text{P}}_2=\;{(50) {\text{g}}(100)}/{2}$ $\;\;{ {\text{P}}_1}/{ {\text{P}}_2}=\;\frac{600}{250}=\;\frac{12}{5}=\;\frac{3 \times 4}{4+1}$ $\;\;{ {\text{P}}_1}/{ {\text{P}}_2}=\;{3 {\sqrt{16}}}/{{\sqrt{16}}+1}$ $\; {\text{x}}=16$

Kinetic energy, $K E=\;\frac{P^2}{2 m}$ $\;\;{ {\text{k}}_1}/{ {\text{k}}_2}=\;{ {\text{m}}_2}/{ {\text{m}}_1}$ $\;\;\frac{16}{9}=\;{ {\text{m}}_2}/{ {\text{m}}_1}$ $\;\;{ {\text{m}}_1}/{ {\text{m}}_2}=\;\frac{9}{16}$

$\;{K E_{\;\text{POP }\;}}/{K E_{\;\text{top }\;}}=\;\frac{\;\frac{1}{2} M(u)^2}{\frac{1}{2} M (u \cos 30^{\circ}) ^2}=\;\frac{4}{3}$

$\; {\text{F}}- {\text{mg}} sin \;30= {\text{ma}}$ $\; {\text{F}}=5 \times 1+25=30 {\text{N}}$ $\; {\text{V}}= {\text{u}}+ {\text{at}}=0+1 \times 10=10$ $\; {\text{P}}= {\text{FV}}=30 \times 10=300 {\text{watt}}$

Work done by a force is given by $W = \vec{F} \cdot \vec{d} = Fd \cos \theta$, where $\theta$ is the angle between the force vector $\vec{F}$ and the displacement vector $\vec{d}$. Work is positive if the force has a component in the direction of displacement and negative if it has a component opposite to the displacement.

(A) The man applies an upward force, and the bucket's displacement is upward. The angle $\theta = 0^\circ$, so the work done is positive. This statement is incorrect.

(B) Gravitational force acts downward, while the bucket's displacement is upward. The angle $\theta = 180^\circ$, so the work done by gravity is negative ($W_g < 0$). This statement is correct.

(C) Frictional force always opposes motion. As the body slides down, friction acts up the incline, opposite to the displacement. The angle $\theta = 180^\circ$, so the work done by friction is negative. This statement is incorrect.

(D) For an object moving with uniform velocity, the net force is zero, meaning the applied force equals the friction force in magnitude. The work done by the applied force is $W_{app} = F_{app}d \cos(0^\circ) > 0$, as the force is in the direction of motion. This statement is incorrect.

(E) Air resistance is a dissipative force that always opposes the velocity of the pendulum bob. Therefore, the angle between the force of air resistance and the displacement is always $180^\circ$, making the work done negative. This statement is correct.

Thus, the correct statements are (B) and (E).

$\; {\text{U}}_{ {\text{i}}}=\;\frac{1}{2} {\text{kx}}_0^2$ $\; {\text{K}}_{ {\text{i}}}=0$ $\; {\text{U}}_{ {\text{f}}}=\;\frac{1}{2} {\text{k}} (\;{ {\text{x}}_0}/{2}) ^2$ $\; {\text{K}}_{ {\text{f}}}=0.25 {\text{J}}$ $\;\;\frac{1}{2} {\text{kx}} {\text{x}}_0^{2+}0=\;\frac{1}{2} {\text{k}} \;{ {\text{x}}_0^2}/{4}+0.25$ $\;\;\frac{1}{2} {\text{kx}} {\text{x}}_0^2 \;\frac{3}{4}=\;\frac{1}{4}$ $\;\;\frac{1}{2} {\text{k}} \;\frac{3}{100}=1 \Rightarrow {\text{k}}=\;\frac{200}{3} {\text{N}} / {\text{m}}$ $\;=67 {\text{N}} / {\text{m}}$