Work Power and Energy

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
$\frac {d K}{d t} = \frac {1}{2} m 2 v \frac {d v}{d t} = v m \frac {d v}{d t} = v \frac {d p}{d t} \quad \left(\because m \frac {d v}{d t} = \frac {d p}{d t}\right)$
$\frac {d K}{d t} = v F$
$\left(\because \text { Force, } F = \frac {d p}{d t}\right)$
$F = \frac {1}{v} \cdot \frac {d K}{d t} = \frac {1}{3} \times 9.6 = 3.2 \mathrm{N}$

$\because$ 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
$\left(\because v = \frac {d}{t}\right)$
$P = 1 1 0 0 0 \times 0. 5 = 5 5 0 0 \mathrm{W} = 5. 5 \mathrm{kW}$

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,
$\begin{array}{c} \mu m g s = m g h \\ h = \mu s \end{array}$

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,
$a _ {T} = \frac {8 \times 1 0 ^ {- 4}}{1 0 ^ {- 2} (2 \pi r \times 2)} = \frac {2 \times 1 0 ^ {- 2}}{\pi \times 6.4 \times 1 0 ^ {- 2}}$
$\simeq 0.1 \mathrm{ms} ^ {- 2}$

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
⇒ $\frac {1}{2} m v ^ {2} = \frac {3 m g l}{8} \Rightarrow v = \sqrt {\frac {3 g l}{4}}$