Work Power and Energy

$W = \int_ {y _ {1}} ^ {y _ {2}} F d y$
$\Rightarrow W = 2 0 [ y ] _ {0} ^ {1} + 1 0 \left[ \frac {y ^ {2}}{2} \right] _ {0} ^ {1}$
$\Rightarrow W=25J$

Work done by a force $F$ , which is variable in nature in moving a particle from $y_{1}$ to $y_{2}$ is given by $W = \int_{y_1}^{y_2} F \cdot dy \quad \ldots$ (i) Given, force, $F = 20 + 10y$ , $y_{1} = 0$ and $y_{2} = 1\mathrm{m}$ Substituting the given values in Eq. (i), we get
$\begin{array}{c} W = \int_ {0} ^ {1} (2 0 + 1 0 y) d y = \left[ 2 0 y + \frac {1 0 y ^ {2}}{2} \right] _ {0} ^ {1} \\ = 2 0 (1 - 0) + 5 (1 - 0) ^ {2} = 2 5 \mathrm{J} \end{array}$
∴ Work done by the force will be 25 J.

In stretching a wire, the work done against internal restoring force is stored as elastic potential energy in the wire and is given by
$U = W = \frac {1}{2} \times \mathrm{Force} (F) \times$
Elongation (l)
$= \frac {1}{2} F l = \frac {1}{2} \times M g \times l = \frac {1}{2} M g l$

The area under the
force-displacement curve gives the amount of work done.
From work-energy theorem, $W = \Delta KE$ ...(i)
$\therefore \mathrm{At} \quad x = 8 \mathrm{m}, W = \mathrm{AreaABDO} + \mathrm{AreaCEFD}$
$= 2 0 \times 5 + 1 0 \times 3 = 1 3 0 \mathrm{J}$

Given, $m = 500\mathrm{g} = 500\times 10^{-3}\mathrm{kg}$
Using Eq. (i), we get
$\Rightarrow 130 = \frac{1}{2} mv^{2} = \frac{1}{2}\times 500\times 10^{-3}\times v^{2}$
$\Rightarrow v = 2\sqrt{130} = 22.8 \, \mathrm{ms}^{-1} \approx 23 \, \mathrm{ms}^{-1}$
At $x = 12\mathrm{m},$
$W = \text { Area } A B D O + \text { Area } C E F D$
+ Area FGHIJ + Area KLMJ
$W = 2 0 \times 5 + 1 0 \times 3 + \left(- 2 0 \times 2 + \frac {1}{2} \times (- 5) \times 2\right)\tag{+ 10 × 2}$
$(\because \text { Area } F G H I J = \text { Area } F G I J + \text { Area } G H I)$
$= 1 0 0 + 3 0 - 4 0 - 5 + 2 0 = 1 0 5 \mathrm{J}$
Using Eq. (i), we get
∴ $105=\frac{1}{2}\times500\times10^{-3}\times v^{2}$
$\Rightarrow v = 2\sqrt{105} \simeq 20.6 \, \mathrm{ms}^{-1}$

If x be the compression in the spring, when a block of 0.25 kg is released.
From law of conservation of energy,
Potential energy of spring = Potential energy of block
$\frac {1}{2} k x ^ {2} = m g x$
where, k is force constant of spring.
$k x = 2 m g = 2 \times 0.2 5 g$
$k x = 0.5 g$
∴Force applied by the spring on the block, F = kx = 0.5 g
From free body diagram,
∴ Maximum force by system on the floor,
$(\because g = 10 \mathrm{~ms}^{-2})$
$N = F + 2 g = 0.5 g + 2 g = 2.5 g$
$(g=10ms^{-2})$ 25N

Given, rate of decrease in kinetic energy,
$\frac {d K}{d t} = 9.6 \mathrm{Js} ^ {- 1}$
When speed, $v = 3 \, ms^{-1}$ , then force F = ?
Kinetic energy, $K = \frac{1}{2} mv^{2}$
Differentiating above equation w.r.t. t, we get
$\frac {d K}{d t} = \frac {1}{2} m 2 v \frac {d v}{d t} = v m \frac {d v}{d t} = v \frac {d p}{d t} \quad \left(\because m \frac {d v}{d t} = \frac {d p}{d t}\right)$
$\frac {d K}{d t} = v F$
$\left(\because \text { Force, } F = \frac {d p}{d t}\right)$
$F = \frac {1}{v} \cdot \frac {d K}{d t} = \frac {1}{3} \times 9.6 = 3.2 \mathrm{N}$

To complete a vertical circle, speed at A should be $v_{A} = \sqrt{5gR}$ using energy conservation $mgh = \frac{1}{2} mv_{A}^{2}$ $h = \frac{1}{2}\frac{v_{A}^{2}}{g} = \frac{1}{2}\frac{5g}{g}\frac{D}{2}\quad \left(R = \frac{D}{2}\right)$ $h = \frac{5D}{4}$

$F = -\frac{\partial u}{\partial r} = \frac{K}{r^3}$
Since it is performing circular motion $F=\frac{mv^{2}}{r}=\frac{K}{r^{3}}$ $mv^{2}=\frac{K}{r^{2}}$ $\Rightarrow K.E.=\frac{1}{2}mv^{2}=\frac{K}{2r^{2}}$ Total energy = P.E. + K.E. $=-\frac{K}{2r^{2}}+\frac{K}{2r^{2}}=Zero$

By conservation of linear momentum
$m v = 4 m v ^ {\prime} \Rightarrow v ^ {\prime} = \frac {v}{4}$
Coefficient of restitution (e) = $\frac{Velocity\ of\ separation}{Velocity\ of\ approach}$
$= \frac {\frac {v}{4} - 0}{v - 0} = \frac {1}{4} = 0. 2 5$

initial

$\frac {1}{2} m v _ {1} ^ {2} + \frac {1}{2} m v _ {2} ^ {2} = \frac {3}{2} \left(\frac {1}{2} m v _ {0} ^ {2}\right)$
$\Rightarrow v _ {1} ^ {2} + v _ {2} ^ {2} = \frac {3}{2} v _ {0} ^ {2}$
from momentum conservation
$m v _ {0} = m (v _ {1} + v _ {2})$....(ii)
$(v _ {1} + v _ {2}) ^ {2} = v _ {0} ^ {2} \Rightarrow v _ {1} ^ {2} + v _ {2} ^ {2} + 2 v _ {1} v _ {2} = v _ {0} ^ {2}$
$2 v _ {1} v _ {2} = - \frac {v _ {0} ^ {2}}{2}$
$(v _ {1} - v _ {2}) ^ {2} = v _ {1} ^ {2} + v _ {2} ^ {2} - 2 v _ {1} v _ {2} = \frac {3}{2} v _ {0} ^ {2} + \frac {v _ {0} ^ {2}}{2}$
$v _ {1} - v _ {2} = \sqrt {2} v _ {0}$

Let initial speed of neutron is $v_{0}$ and kinetic energy is K. $1^{st}$ collision:

By momentum conservation
$m v _ {0} = m v _ {1} + 2 m v _ {2} \Rightarrow v _ {1} + 2 v _ {2} = v _ {0}$
by $e = 1$ $v_{2} - v_{1} = v_{0}$
$\Rightarrow \quad v _ {2} = \frac {2 v _ {0}}{3}; v _ {1} = - \frac {v _ {0}}{3}$
$\text { fractional loss } = \frac {\frac {1}{2} m v _ {0} ^ {2} - \frac {1}{2} m \left(\frac {v _ {0}}{3}\right) ^ {2}}{\frac {1}{2} m v _ {0} ^ {2}}$
$\Rightarrow P _ {d} = \frac {8}{9} \approx . 89$
$2^{\mathrm{nd}}$ collision:

by momentum conservation
$m v _ {0} = m v _ {1} + 1 2 m v _ {2}$
$\Rightarrow v _ {1} + 1 2 v _ {2} = v _ {0}$
by $e = 1$
$v _ {2} - v _ {1} = v _ {0}$
$v _ {2} = \frac {2 v _ {0}}{1 3}; v _ {1} = \frac {- 1 1 v _ {0}}{1 3}$
Now fraction loss of energy

$\because$ Efficiency $= \frac{W_o}{W_i} \times 100$
Given, $W_{o} = 4000 \mathrm{~J}$ and $W_{i} = 4000 + 1000 = 5000 \mathrm{~J}$
$\because$ Efficiency $(\eta) = \frac{4000}{5000}\times 100 = 80\%$

Relation between kinetic energy and momentum is
$p _ {1} = \sqrt {2 m K _ {1}}$
∵ Kinetic energy is increased by 4 times, then $K_{2} = 4K_{1}$
Hence, $p_2 = \sqrt{2mK_2} = \sqrt{2m(4K_1)}$
$p _ {2} = 2 \sqrt {2 m K _ {1}} \mathrm{or} p _ {2} = 2 p _ {1}$

Net force required to lift a hook and load,
$F _ {\mathrm{net}} = 1 0 0 0 + 1 0 0 0 0 = 1 1 0 0 0 \mathrm{N}$
Power required to lift the hook, $P = \frac{W}{t}$
As,
$W = F _ {\mathrm{net}} d$
∴ $P = \frac {F _ {\mathrm{net}} d}{t} = F _ {\mathrm{net}} \left(\frac {d}{t}\right) = F _ {\mathrm{net}} v$
or
$\left(\because v = \frac {d}{t}\right)$
$P = 1 1 0 0 0 \times 0. 5 = 5 5 0 0 \mathrm{W} = 5. 5 \mathrm{kW}$

Initially

Finally
Momentum Conservation
$m v = m v _ {A} + \frac {m}{2} v _ {B}$
... (i)
$e = \frac {v _ {B} - v _ {A}}{v} = 1$....(ii)
$2 (v _ {B} - v _ {A}) = 2 v _ {A} + v _ {B}$
$2 v _ {B} - v _ {B} = 4 v _ {A}$
$v _ {B} = 4 v _ {A}$
$\frac {\lambda_ {A}}{\lambda_ {B}} = \frac {m _ {B} v _ {B}}{m _ {A} v _ {A}} = \frac {m / 2}{m} \times 4 = 2$

$w_{g} = mgh = 10^{-3} \times 10 \times 10^{3} = 10J$
$w _ {n e t} = \Delta K E = \frac {1}{2} \times 1 0 ^ {- 3} \times 5 0 \times 5 0 = \frac {2 . 5}{2} J = 1. 2 5 J$
By work-energy theorem,
$w _ {g} + w _ {r e s} = w _ {n e t}$
$1 0 + w _ {r e s} = 1.25$
$\therefore w _ {r e s} = - 8.7 5 J$

$F = -Kv^{2}$
$m\frac{dv}{dt} = -kv^{2}$
$\int_ {v _ {0}} ^ {v} v ^ {- 2} d v = \int_ {0} ^ {t} - \frac {k}{m} d t$
After 10 s, $KE = \frac{1}{2} mv^2 = \frac{1}{8} mv_0^2 \Rightarrow v = \frac{v_0}{2}$
$\left[ - \frac {1}{v} \right] _ {v _ {0}} ^ {v _ {0} / 2} = - \frac {k}{m} t$
$\left(\frac {2}{v _ {0}} - \frac {1}{v _ {0}}\right) = \frac {k}{m} t \Rightarrow k = \frac {m}{v _ {0} t} = \frac {1 0 ^ {- 2}}{1 0 \times 1 0}$
$k = 1 0 ^ {- 4} k g m ^ {- 1}$

$a = \frac{F}{m} = \frac{6t}{1} = 6t$
$\frac {d v}{d t} = 6 t$
$d v = 6 t d t$
$\int_ {0} ^ {v} d v = 6 \int_ {0} ^ {v} t d t$
$v = 6 \left[ \frac {t ^ {2}}{2} \right] _ {0} ^ {1} = 3$
$W = \Delta K E = K _ {F} - K _ {i}$
$= \frac {1}{2} (1) (3) ^ {2} = 4.5 J$

When the spring is cut into pieces, they will have the new force constant. The spring is divided into 1:2:3 ratio.
When the pieces are connected in series, the resultant force constant will be given by
$\frac {1}{k ^ {\prime}} = \frac {1}{k _ {1}} + \frac {1}{k _ {2}} + \frac {1}{k _ {3}}$
$\frac {1}{k ^ {\prime}} = \frac {1}{x} + \frac {1}{2 x} + \frac {1}{3 x} \Rightarrow k ^ {\prime} = \frac {6 x}{1 1}$
In parallel, the net force constant,
$k ^ {\prime \prime} = x + 2 x + 3 x = 6 x$
The required ratio $\frac{k'}{k''}=\frac{6x/11}{6x}=1:11$

Given, $\mathbf{F} = -k(y\hat{\mathbf{i}} + x\hat{\mathbf{j}})$
As, work done, $W = \int \mathbf{F} \cdot d\mathbf{r}$
So,
$W = - k \int (\hat {y \mathbf {i}} + \hat {\mathbf {x j}}) \cdot (d x \hat {\mathbf {i}} + d y \hat {\mathbf {j}})$
⇒ $W = - k \int (y d x + x d y)$

$\Rightarrow \quad W = - k \int_ {(0, 0)} ^ {(a, a)} d (x y) \quad [ \because y d x + x d y = \int d (x y) ]$
Hence, $W = -k(xy)_{(0,0)}^{(a,a)} = -ka^2$

As we know, $k = \frac{F}{l} \Rightarrow k \propto \frac{1}{l}$
⇒ $\frac {k _ {2}}{k _ {1}} = \frac {l _ {1}}{l _ {2}} = \frac {1}{2}$
$k _ {1} = 2 k, k _ {2} = k$
In series, $\frac{1}{k'} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{1}{2k} + \frac{1}{k} = \frac{3}{2k}$
$k ^ {\prime} = \frac {2 k}{3}$
So, both Assertion and Reason are incorrect.

Free body diagram of block is as shown below.

Now, from law of conservation of energy, $W = \Delta K$
$\Rightarrow W _ {F} + W _ {s p} = \frac {1}{2} m v ^ {2}$
$\Rightarrow F \cdot x - \frac {1}{2} k x ^ {2} = \frac {1}{2} m v ^ {2}$

Given, $m = 70 \, kg$ , $g = 10 \, ms^{-2}$ and $h = 12 \, m$
In going up and down once,
Number of k-cal burnt = $\left(mgh + \frac{mgh}{2}\right)$ $=\frac{3}{2}mgh=\frac{3}{2}\times\frac{70\times10\times12}{4.2\times1000}$ = 3 k-cal
Total number of k-cal to be burnt to loose 7 kg of weight $= 7 \times 9000 = 63000$ k-cal
∴Number of times the person has to go up and down the stairs
$= \frac {6 3 0 0 0}{3} = 2 1 0 0 0 = 2 1 \times 1 0 ^ {3} \mathrm{times}$

According to question, the given situation is shown in the figure below.

For a sufficiently safe-horizontal displacement, $\Delta s$ can be considered straight. If the corresponding length of path element is $\Delta L$ , the frictional force is given by
$\Delta F = \mu mg(\Delta L\cos \theta)$
From $\Delta OPQ$ , $\cos \theta = \frac{\Delta s}{\Delta L}$
So, $\Delta F = \mu mg\Delta L\left(\frac{\Delta s}{\Delta L}\right) = \mu mg\Delta s$
On adding up, we find that along the whole path, the total work done by the friction force is $\mu mgs$ . By law of conservation of energy, this must equals to the decrease in potential energy of skier.
Hence,
$\begin{array}{c} \mu m g s = m g h \\ h = \mu s \end{array}$

$\vec{S} = \vec{r}_f - \vec{r}_i = (4\hat{j} + 3\hat{k}) - (-2\hat{\mathrm{i}} + 5\hat{\mathrm{j}}) = 2\hat{\mathrm{i}} - \hat{\mathrm{j}} + 3\hat{k}$ and $\vec{F} = 4\hat{i} + 3\hat{j}$ $\omega = \vec{F}. \vec{S} = (4\hat{i} + 3\hat{j}). (2\hat{i} - \hat{j} + 3\hat{k}) = 8 - 3 = 5J$