Work Power and Energy

Given that $\frac{\mu mgh}{\tan\theta} = mgh - \frac{\mu mgh}{\tan\theta}$
$\frac {2 \mu}{\tan \theta} = 1 \Rightarrow \mu = \frac {\tan \theta}{2}$
$\mu = 0.2 9$
$x = \frac {h}{\tan \theta} = 2 \sqrt {3} \simeq 3. 5 m.$

During the collision, apply momentum conservation
(0.01)(400) + 0 = (2)V + (0.01)V' ... (i)
where V = $\sqrt{2gh} = \sqrt{2 \times 10 \times 0.1} = \sqrt{2}$ from (i) $4 = 2 \times \sqrt{2} + 0.01V'$ solving V' = 120m/sec.

Let $m$ mass of fat is used $m(3.8 \times 10^{7})\frac{1}{5} = 10(9.8)(1)(1000)$ $m = \frac{9.8 \times 5}{3.8 \times 10^{3}} = 12.89 \times 10^{-3}kg.$

$\vec{F} = 2t\hat{i} + 3t^2\hat{j}$
$m \frac {d \vec {v}}{d t} = 2 t \hat {i} + 3 t ^ {2} \hat {j} (m = 1 k g)$
$\Rightarrow \int_ {0} ^ {\vec {v}} d \vec {v} = \int_ {0} ^ {t} (2 t \hat {i} + 3 t ^ {2} \hat {j}) d t \Rightarrow \vec {v} = t ^ {2} \hat {i} + t ^ {3} \hat {j}$
$\mathrm{Power} = \vec {F}. \vec {v} = (2 t ^ {3} + 3 t ^ {5}) W.$

Mass of balls are same and the collision is perfectly elastic, so their velocity will be interchanged. i.e. $V_{A} = -0.3m / s, V_{B} = 0.5m / s$ .

According to question, a body of mass 1 kg begins to move under the action of time dependent force,
$\begin{array}{c} \mathbf {F} = (2 t \hat {\mathbf {i}} + 3 t ^ {2} \hat {\mathbf {j}}) \mathrm{N} \\ m \frac {d v}{d t} = 2 t \hat {\mathbf {i}} + 3 t ^ {2} \hat {\mathbf {j}} \\ \int d v = \int (2 t \hat {\mathbf {i}} + 3 t ^ {2} \hat {\mathbf {j}}) d t \\ \mathbf {v} = t ^ {2} \hat {\mathbf {i}} + t ^ {3} \hat {\mathbf {j}} \end{array}$
$(\because m = 1 \mathrm{kg})$
∴ Power developed by the force at the time t will be given by
$\begin{array}{r l} & P = \mathbf {F} \cdot \mathbf {v} = (2 t \hat {\mathbf {i}} + 3 t ^ {2} \hat {\mathbf {j}}) \cdot (t ^ {2} \hat {\mathbf {i}} + t ^ {3} \hat {\mathbf {j}}) \\ & \quad = (2 t \cdot t ^ {2} + 3 t ^ {2} \cdot t ^ {3}) \\ & P = (2 t ^ {3} + 3 t ^ {5}) \mathrm{W} \end{array}$

From work-energy theorem,
$\begin{array}{l} W _ {\mathrm{tangential}} = \Delta \mathrm{KE} = K _ {f} - K _ {i} = K _ {f} - 0 \\ (m a _ {T}) \times S = 8 \times 1 0 ^ {- 4} \end{array}$
Here,
$a _ {T} = \frac {8 \times 1 0 ^ {- 4}}{1 0 ^ {- 2} (2 \pi r \times 2)} = \frac {2 \times 1 0 ^ {- 2}}{\pi \times 6.4 \times 1 0 ^ {- 2}}$
$\simeq 0.1 \mathrm{ms} ^ {- 2}$

$a = \frac{0.1x}{10} = 0.01x = V\frac{dV}{dx}$ So, $\int_{v_1}^{v_2} v dV = \int_{20}^{30} \frac{x}{100} dx$ $-\frac{V^2}{2}\bigg|_{v_1}^{v_2} = \frac{x^2}{200}\bigg|_{20}^{30} = \frac{30 \times 30}{200} - \frac{20 \times 20}{200} = 4.5 - 2 = 2.5$ $\frac{1}{2} m(v_2^2 - v_1^2) = 10 \times 2.5J = -25J$ $\frac{1}{2} mv_2^2 = \frac{1}{2} mv_1^2 - 25 = \frac{1}{2} \times 10 \times 10 \times 10 - 25$ $= 500 - 25J = 475J$ .

$\frac{dw}{dt} = P$
$w = P t = \frac {1}{2} m V ^ {2}$
So $\sqrt{\frac{2Pt}{m}} = V$
Hence $a = \frac{dV}{dt} = \sqrt{\frac{2P}{m}}\cdot \frac{1}{2\sqrt{t}}$
so from = ma = $\sqrt{\frac{2Pm^{2}}{m}}$ . $\frac{1}{2\sqrt{t}} = \sqrt{\frac{Pm}{2t}}$ .

For collision $\vec{v}_{B / A}$ should be along $\overrightarrow{B\to A}$ ( $\vec{r}_{A / B}$ )
$\mathrm{So}, \frac {\stackrel {\rightarrow} {v _ {2}} - \stackrel {\rightarrow} {v _ {1}}}{| v _ {2} - v _ {1} |} = \frac {\stackrel {\rightarrow} {r _ {1}} - \stackrel {\rightarrow} {r _ {2}}}{| r _ {1} - r _ {2} |}$

$E_{1} = \frac{1}{2} m(2v)^{2} + \frac{1}{2} 2m.v^{2} = \frac{1}{2} m.4v^{2} + mv^{2}$
$= 2 m v ^ {2} + m v ^ {2} = 3 m v ^ {2}$
After Collision
3mV = $\sqrt{2}$ . 2mv
$V = \frac {2 \sqrt {2} v}{3}$
$E _ {2} = \frac {1}{2} 3 m. \left(\frac {2 \sqrt {2} v}{3}\right) ^ {2} = \frac {3}{2} m. \frac {8 v ^ {2}}{9} = \frac {4}{3} m v ^ {2}$
$E _ {1} - E _ {2} = 3 m v ^ {2} - \frac {4}{3} m v ^ {2} = \frac {9 m v ^ {2} - 4 m v ^ {2}}{3} = \frac {5}{3} m v ^ {2}$
$\therefore \frac {E _ {1} - E _ {2}}{E _ {1}} = \frac {\frac {5}{3} m v ^ {2}}{3 m v ^ {2}} = \frac {5}{9}$
Percentage loss= $\frac{5}{9}\times100=55.6\approx56.$

$\frac{1}{2} mv_1^2 +\frac{1}{2} mv_2^2 = \frac{1}{2} m_1u_1^2 +\frac{1}{2} m_2u_2^2 -\varepsilon$
By energy conservation.

From work-energy theorem,
Work done = Change in kinetic energy $\Rightarrow W = K_{f} - K_{i}$
$\begin{array}{r l} \Rightarrow & K _ {f} = W + K _ {i} = \int_ {x _ {1}} ^ {x _ {2}} F d x + \frac {1}{2} m v ^ {2} \\ & = \int_ {2 0} ^ {3 0} - 0.1 x d x + \frac {1}{2} \times 1 0 \times 1 0 ^ {2} \\ & = - 0.1 \left[ \frac {x ^ {2}}{2} \right] _ {2 0} ^ {3 0} + 5 0 0 = - 0.0 5 [ 3 0 ^ {2} - 2 0 ^ {2} ] + 5 0 0 \\ & = - 0.0 5 [ 9 0 0 - 4 0 0 ] + 5 0 0 \end{array}$
$\Rightarrow K _ {f} = - 2 5 + 5 0 0 = 4 7 5 \mathrm{J}$

As, the machine delivers a constant power.
${\frac {v ^ {2}}{2}} = {\frac {k}{m}} t \Rightarrow v = \sqrt {{\frac {2 k}{m}} t}$
Now, force on the particle is given by
$F = m \frac {d v}{d t} = m \frac {d}{d t} \left(\frac {2 k t}{m}\right) ^ {\frac {1}{2}}$
$= \sqrt {2 k m} \left(\frac {1}{2} t ^ {- \frac {1}{2}}\right) = \sqrt {\frac {m k}{2}} \cdot t ^ {- 1 / 2}$

Given, force, $F = 10 + 0.5x = 10 + \frac{1}{2}x$
Let during small displacement, the work done by the force is $dW = Fdx$ .
So, work done during displacement from x = 0 to x = 2 is
$\begin{array}{r l} W & = \int_ {0} ^ {x} d W = \int_ {0} ^ {2} F d x = \int_ {0} ^ {2} \left(1 0 + \frac {1}{2} x\right) d x \\ & = 1 0 [ x ] _ {0} ^ {2} + \left[ \frac {x ^ {2}}{4} \right] _ {0} ^ {2} = 2 1 \mathrm{J} \end{array}$

Given, $m = 11.7 \, kg$ , $s = 4.65 \, m$ , $h = 2.86 \, m$
From the figure, $\sin \theta = \frac{h}{s} = \frac{2.86}{4.65} = 0.615$
Also, component of mg along the incline is $mg \sin \theta$ .
∴ Work done by this force = mg sin θ · s
= 11.7 × 9.8 × 0.615 × 4.65
= 11.7 × 9.8 × 2.86 = 327.9 ≈ 328 J

Given, mass of elevator = 500 kg
$\mathrm{Velocity} = 0. 2 0 \mathrm{ms} ^ {- 1}$
Weight of elevator = 500 × 9.8 = F
Now, power, $P = Fv = 500 \times 9.8 \times 0.20 = 980 \mathrm{~W}$
Therefore, hp-rating of motor = $\frac{1}{746} \times 980 = 1.31hp$

Let the kinetic energy is K and masses of bodies are $m_{1}$ and $m_{2}$ .
For first body, $p_1 = \sqrt{2m_1K}$
$p _ {1} ^ {2} = 2 m _ {1} K$...(i)
Similarly, for second body, $p_{2}^{2}=2m_{2}K$
... (ii)
Dividing Eq. (i) by Eq. (ii), we get
⇒ $\begin{array}{l} \frac {p _ {1} ^ {2}}{p _ {2} ^ {2}} = \frac {2 m _ {1} K}{2 m _ {2} K}, \frac {p _ {1}}{p _ {2}} = \sqrt {\frac {m _ {1}}{m _ {2}}} \\ \frac {p _ {1}}{p _ {2}} = \text {square root of masses.} \end{array}$

Given, $m_{1}=1$ kg and $m_{2}=2$ kg
Let kinetic energies of bodies be $K_{1}$ and $K_{2}$ .
For first body,
$K _ {1} = F s _ {1}$
For second body, ... (i)
$K _ {2} = F s _ {2}$...(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac {s _ {1}}{s _ {2}} = \frac {K _ {1}}{K _ {2}} \Rightarrow \frac {s _ {1}}{s _ {2}} = \frac {m _ {1} v _ {1} ^ {2}}{m _ {2} v _ {2} ^ {2}}$
Given,
$v _ {1} = v _ {2}, \mathrm{so} \frac {s _ {1}}{s _ {2}} = \frac {m _ {1}}{m _ {2}}, \frac {s _ {1}}{s _ {2}} = \frac {1}{2}$

From $v = u + at$ , $v_{1} = 0 + at_{1}$
$F = m a = \frac {m v _ {1}}{t _ {1}}$
Velocity acquired in $t$ second $= at \Rightarrow v = \frac{v_1}{t_1} t$
$\left(\because a = \frac {v _ {1}}{t _ {1}}\right)$
$\mathrm{Power} = Fv = \left(\frac{mv_1}{t_1}\right)\left(\frac{v_1}{t_1} t\right) = \left(\frac{mv_1^2}{t_1^2}\right)t$

Consider the diagram. Let the mass of chain be m.
∴ Mass per unit length = $\frac{m}{l}$
Consider an elementary length of the chain at a depth x below the table. The potential energy of their parts is given by

$d U = - \frac {m}{l} g x d x$
Now, potential energy of hanging part,
$U _ {1} = \int d U = - \int_ {0} ^ {\frac {l}{2}} \frac {m}{l} g x d x = - \frac {m}{l} g \left[ \frac {x ^ {2}}{2} \right] _ {0} ^ {\frac {l}{2}} = - \frac {m g l}{8}$
When chain is completely leaves the table, then potential energy,
$U _ {2} = - \int_ {0} ^ {l} \frac {m}{l} g x d x = \frac {m g}{2 l} [ x ^ {2} ] _ {0} ^ {l} = \frac {- m g l}{2}$
Now, loss in potential energy = $\frac{-mgl}{8} + \frac{mgl}{2}$
From law of conservation of energy,
Loss in potential energy = Gain in kinetic energy
⇒ $\frac {1}{2} m v ^ {2} = \frac {3 m g l}{8} \Rightarrow v = \sqrt {\frac {3 g l}{4}}$

Given, $m = 4 \, kg$ , $p = 6 \, N \cdot s$
We know that, kinetic energy of a body,
$K = \frac {p ^ {2}}{2 m}, K = \frac {(6) ^ {2}}{2 \times 4} \Rightarrow K = \frac {3 6}{2 \times 4} = 4. 5 \mathrm{J}$

Given, $m = 3.90 \, kg$ , $v = 1.20 \, ms^{-1}$
and k for spring = 135Nm $^{-1}$
From law of conservation of energy,
$\begin{array}{r l} \frac {1}{2} m v ^ {2} & = \frac {1}{2} k x ^ {2} \Rightarrow x = \sqrt {\frac {m v ^ {2}}{k}} \\ & = \sqrt {\frac {3.9 0 \times 1.2 0 \times 1.2 0}{1 3 5}} \\ & = 0.2 0 4 \mathrm{m} \end{array}$

Work done in stretching the rubber band $W = \int_{0}^{L}(ax + bx^{2})dx = \frac{aL^2}{2} +\frac{bL^3}{3}$ .

Initial momentum $= P_{i} = 0$
Final momentum $P_{f} = 0 = mv\hat{i} + mv\hat{j} +\bar{P}_{3}$
$\Rightarrow P _ {3} = m v \sqrt {2}$
Total KE= $\frac{P_{3}^{2}}{2\times2m}+\frac{1}{2}mv^{2}+\frac{1}{2}mv^{2}$
$= \frac {2 m ^ {2} v ^ {2}}{4 m} + m v ^ {2} = \frac {3 m v ^ {2}}{2}.$