Work Power And Energy Practice Questions

The potential energy stored in a spring of constant $k$ supporting a mass $m$ under gravity is given by $U = \frac{(mg)^2}{2k}$.
Given $m_A = 100\text{ g}$ and $m_B = 200\text{ g}$, the energy in spring $A$ is $E = \frac{(m_A g)^2}{2k_A}$ and the energy in spring $B$ is $E_B = \frac{(m_B g)^2}{2k_B}$.
Taking the ratio:
$\frac{E_B}{E} = \left( \frac{m_B}{m_A} \right)^2 \cdot \frac{k_A}{k_B} = \left( \frac{200}{100} \right)^2 \cdot \frac{k_A}{\frac{4}{3}k_A}$
Substituting the given values:
$\frac{E_B}{E} = 2^2 \cdot \frac{3}{4} = 4 \cdot \frac{3}{4} = 3 \implies E_B = 3E$

The kinetic energy at the projection point $A$ is $K_A = K$. At the highest point $B$, the vertical component of velocity is zero, so the kinetic energy is $K_B = K \cos^2(60^{\circ}) = K(\frac{1}{2})^2 = \frac{K}{4}$. Since points $A$ and $C$ are at the same horizontal level, the speed at $C$ equals the speed at $A$, thus $K_C = K_A = K$. The difference in kinetic energies at points $B$ and $C$ is $|K_C - K_B| = |K - \frac{K}{4}| = \frac{3K}{4}$. The required ratio is $\frac{3K/4}{K} = \frac{3}{4}$, which is $3:4$.

The relationship between force $F$ and potential energy $V$ is given by $F = -\frac{dV}{dx}$. The magnitude of the force $|F|$ is equal to the absolute value of the slope of the $V-x$ graph, $| \frac{dV}{dx} |$. Calculating the slopes for each region:

1. Region AB: Slope $= \frac{1-0}{1-0} = 1$, so $|F_{AB}| = 1$.
2. Region BC: Slope $= \frac{4-1}{2-1} = 3$, so $|F_{BC}| = 3$.
3. Region CD: Slope $= \frac{4-4}{4-2} = 0$, so $|F_{CD}| = 0$.
4. Region DE: Slope $= \frac{3-4}{6-4} = -0.5$, so $|F_{DE}| = |-0.5| = 0.5$.

Comparing the magnitudes, we get $3 > 1 > 0.5 > 0$, which corresponds to $F_{BC} > F_{AB} > F_{DE} > F_{CD}$.

Bead $Q$ moves with constant speed $v$ along the horizontal distance $AB = 2R$, so the time taken is:
$t_Q = \frac{2R}{v}$
Bead $P$ slides along the circular path $SB$, gaining kinetic energy due to gravity as it descends, which significantly increases its speed $v_P$ along the path. Although the path length of $P$ is greater than the horizontal displacement, the increase in its horizontal velocity component $v_x = v_P \cos \theta$ is sufficient to make its average horizontal velocity higher than the constant velocity $v$ of bead $Q$. Consequently, bead $P$ reaches point $B$ in less time than bead $Q$. Therefore, the relation is:
$t_P < t_Q$

Given $x(t) = \alpha t^2 + \beta t + \gamma = t^2 + t + 1$, the velocity is $v(t) = \frac{dx}{dt} = 2t + 1$.
At $t=2 \text{ s}$, the initial velocity is $v_i = 2(2) + 1 = 5 \text{ m/s}$.
At $t=3 \text{ s}$, the final velocity is $v_f = 2(3) + 1 = 7 \text{ m/s}$.
By the Work-Energy Theorem, the work done is $W = \Delta K = \frac{1}{2} m (v_f^2 - v_i^2)$.
Substituting the given mass $m = 2 \text{ kg}$ and velocities: $W = \frac{1}{2} \times 2 \times (7^2 - 5^2) = 49 - 25 = 24 \text{ J}$.

Work done by gravity $=m g (2 R-R \cos 60^{\circ})$ $=\;\frac{3 m g R}{2}$Work done by spring $=-\;\frac{1}{2} k (0^{2-}R^2)$ $=\;\frac{1}{2} k R^2$Net work = change in kinetic energyi.e. $\;\frac{3 m g R}{2}+\;\frac{k R^2}{2}=\;\frac{1}{2} m v^2$or $\;\; v^2=3 g R+\;\frac{k R^2}{m}$or $\;\; v=\sqrt{3 g R+\;\frac{k R^2}{m}}$

$\vec{F}=(2 {i}^{∧}+3 {j}^{∧}) N ;$ displacement $\vec{S}=(3 {i}^{∧}+6 {j}^{∧}) m$Acceleration, $\vec{a}=\;\frac{\vec{F}}{m}= (\;\frac{2 {i}^{∧}+3 {j}^{∧}}{4}) m / s ^2$ $\;\vec{S}=\vec{u} t+\;\frac{1}{2} \vec{a} t^2$ $\; \Rightarrow 3 {i}^{∧}+6 {j}^{∧}=4 \vec{u}+\;\frac{1}{2} (\;\frac{2 {i}^{∧}+3 {j}^{∧}}{4}) \cdot 16=4 \vec{u}+4 {i}^{∧}+6 {j}^{∧}$ $\;\text{ or }\; \vec{u}=\;\frac{1}{4}(3 {i}^{∧}+6 {j}^{∧}-4 {i}^{∧}-6 {j}^{∧})$ $\vec{u}=-\;\frac{1}{4} {i}^{∧}$ $\;\;\text{ Average Power }\;=\vec{F} \cdot \;\frac{\vec{S}}{t}$ $\; \;\;\;\; =\;\frac{(2 {i}^{∧}+3 {j}^{∧}) \cdot (3 {i}^{∧}+6 {j}^{∧})}{4} \;\text{ watt }\;$ $\; \;\;\;\; =6 \;\text{ watt }\;$ $\;\text{ Instantaneous Power }\;\;=\vec{F} \cdot \vec{v}$ $\;=\vec{F} \cdot (\vec{u}+\vec{a} t)$ $\;=(2 {i}^{∧}+3 {j}^{∧}) \cdot (-\;\frac{1}{4} {i}^{∧}+2 {i}^{∧}+3 {j}^{∧})$ $\;=(2 {i}^{∧}+3 {j}^{∧}) \cdot (+\;\frac{7}{4} {i}^{∧}+3 {j}^{∧})$ $\;=(3.5+9) \;\text{ watt }\;$ $\;=12.5 \;\text{ watt }\;$

For a block moving at uniform velocity, the net force is zero: $F\cos\theta = \mu N$ and $N = mg - F\sin\theta$.
Substituting the normal force $N$, we get $F\cos\theta = \mu(mg - F\sin\theta)$, which simplifies to $F(\cos\theta + \mu\sin\theta) = \mu mg$.
Rearranging for the horizontal component of the force ($F_h = F\cos\theta$):
$F\cos\theta = \frac{\mu mg \cos\theta}{\cos\theta + \mu\sin\theta} = \frac{\mu mg}{1 + \mu\tan\theta}$
Given $m=25$ kg, $g=9.8$ m/s$^2$, $\mu=0.25$, $\theta=45^{\circ}$, and $d=5$ m, the horizontal force is $F\cos\theta = \frac{0.25 \times 25 \times 9.8}{1 + 0.25 \times 1} = \frac{61.25}{1.25} = 49$ N.
The work done is $W = (F\cos\theta) \times d = 49 \times 5 = 245$ J.

Let speed at projection is $v_0$ at highest point it would be $v_0 \cos \theta$ as $K E \propto V^2$ $KE_{\;\text{Highest }\;}\;=KE_{\;\text{Projection }\;} \cos ^2 \theta$ $\;=\;\frac{1}{4} KE_{\;\text{Projection }\;}$

$\;y=10-x$ $\;w=∫_{0}^{4} x^2(10-x) \text{dx}+∫_{0}^{2} y^2 d y$ $={\frac{10 x^3}{3}-\;\frac{x^4}{4} |}_0 ^{4+}{\frac{y^3}{3} |}_0 ^2$ $\;=\;\frac{640}{3}-\;\frac{256}{4}+\;\frac{8}{3}$ $\;=216 \times 64$ $\;=152 {\text{J}}$

$F=\alpha +\beta x^2$Work done $∫ d w=∫ F \cdot \text{dx}$ $\;\Rightarrow \;\; \Delta w=∫ F \cdot \text{dx}=∫ (\alpha +\beta x^2) \text{dx}$ $\;\Rightarrow \;\; \Delta w={| \alpha x+\;\frac{\beta x^3}{3} |}_0^1=\alpha +\;\frac{\beta }{3}=5$Given $\alpha =1$So, $\;\frac{\beta }{3}=4$ $\Rightarrow \beta =12 N / m^2$

Potential energy can be expressed for conservative force, so it can not be expressed for friction because it is non-conservative force

$\; {\text{PE}}=m g h$ $\; {\text{KE}}=m g(S-h)$ $\;\;\text{ given }\; m g(S-h)=3 m g h$ $\;S=4 h$ $\;h=\;\frac{S}{4}$ $\;U=\sqrt{\;\frac{2 g(3 S)}{4}}=\sqrt{\;\frac{3 g S}{2}}$

Power $=\vec{F} \cdot \vec{V}$and $F=\;\frac{d p}{d t}=$ Rate of change of linear momentum$F=V \cdot \;\frac{d m}{d t}=K_1 V^{\;\frac{3}{2}}, K$ is constantPower $(P)= (K V^{\;\frac{3}{2}}) \cdot (V)$ $=K V^{\;\frac{5}{2}}$So, $P^2 \propto V^5$

According to Pascal's law$\;\frac{F_1}{A_1}\;=\;\frac{F_2}{A_2}$ $F_2\;= (\;\frac{A_2}{A_1}) F_1=\;\frac{1500}{6} \times 100 N$ $W\;=F_2 d_2=\;\frac{1500}{6} \times 100 \times \;\frac{20}{100}$ $\;=5 kJ$

$\;w=0$ $\; \therefore \;\; \vec{F} \cdot \vec{S}=0$ $\;(2 \hat{i}+b \hat{j}+\hat{k}) \cdot (\hat{i}-2 \hat{j}-\hat{k})=0$ $\;2-2 b-1=0$ $\;b=\;\frac{1}{2}$

$m=1000$ gram$F= (2 t {i}^{∧}+3 t^2 {j}^{∧})$So, $\;\frac{d v}{d t}=\;\frac{F}{m}= (2 t {i}^{∧}+3 t^2 {j}^{∧})$So $∫_{0}^{v} d v=∫_{0}^{t} (2 t {i}^{∧}+3 t^2 {j}^{∧}) \text{dt}$So $v= (t {i}^{∧}+t^3 {j}^{∧})$So power $=\vec{F} \cdot \vec{v}=2 t^{2+}3 t^5$

Assertion is correct as central forces are conservative in nature, i.e. work done is independent of path.Reason is true as some forces in mechanics like, friction are non-conservative because work done depends on path and only conservative forces have an associated potential energy.Also, reason does not explain assertion.

${\text{K}}_{ {\text{i}}}=\;\;\frac{1}{2} {\text{m}}(100)^2$ ${\text{K}}_{ {\text{f}}}\;=\;\frac{1}{2} {\text{m}}(40)^2$ $\% \;\text{ loss }\;\;=\;{{| {\text{K}}_{ {\text{f}}}- {\text{K}}_{ {\text{i}}} |}}/{ {\text{K}}_{ {\text{i}}}} \times 100$ $\;=\;{{| \frac{1}{2} {\text{m}}(40)^{2-}\frac{1}{2} {\text{m}}(100)^2 |}}/{\frac{1}{2} {\text{m}}(100)^2} \times 100$ $\;=\;\frac{|1600-100 \times 100|}{100}=84 \%$