(I) Values are: Intelligence, dedication, loving, social and cooperative.
(ii) Human heart rate is about 70 beats /min and scale factor is 2.5.
(iii) Monkey's heart rate = 70 x 2.5 = 175.

Initial Kinetic energy = \(\frac { 1 }{ 2 } { mv }^{ 2 }\)
Final Energy = \(\frac { 1 }{ 2 } m{ \left( \frac { v }{ 2 } \right) }^{ 2 }+\frac { 1 }{ 2 } { kx }^{ 2 }\)
By the principle of conservation of energy,
\(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } \frac { { mv }^{ 2 } }{ 4 } +\frac { 1 }{ 2 } { k }x^{ 2 }\)
\(\therefore \quad K=\frac { 3{ mv }^{ 2 } }{ 4{ x }^{ 2 } } \)

Mass of the fan, m = 2 Kg
Length of the elevator, h = 2 m.
The kinetic energy acquired by the fan during its fall is:
E = mgh = 2 \(\times\) 9.8 \(\times\) 2 = 39.2 J
The amount of heat produced will also be 39.2 J, as in accordance with the law of conservation of energy, whole of the kinetic enery will be converted into heat.

Let the ball fall from a height, h then
K.E. of ball at the time of just striking the ground = P.E. of ball at height h
\(\Rightarrow\) K = mgh
Similarly on rebounding the ball moves to a maximum height h', then K.E. of ball on rebounding
K' = P.E. of ball at a height h' = mgh'
\(\therefore\) Loss of K.E. due to the rebounce K-K' = mgh - mgh' = mg(h-h')
\(=mg\left( h-\frac { 80 }{ 100 } h \right) =mgh\times (0.2)\)
\(\therefore\) Fractional loss in K.E. of ball in each rebounce = \(\frac { K-K' }{ K } =\frac { mgh\times (0.2) }{ mgh } =0.2\)
= 0.2 \(\times\) 100% = 20%

Here,  m = 2 kg, u = 0, F = 5N, t = 10s,
\(\therefore\) Acceleration produced in the body
\(a=\frac { F }{ m } =\frac { 5 }{ 2 } =2.5\) ms^-2
Velocity attained by the body after 10s is
v = u + at = 0 + 2.5 \(\times\) 10 = 25 ms^-1
\(\therefore\) Work done = chage in K.E.
\(=\frac { 1 }{ 2 } { mv }^{ 2 }-\frac { 1 }{ 2 } { mv }^{ 2 }\)
\(=\frac { 1 }{ 2 } \times 2\times 625-0=625\ J\)
\(\therefore\) p = \(\frac { W }{ t } =\frac { 625 }{ 10 } \)
\(\Rightarrow\) p = 62.5 Js^-1 = 62.5 W.

Graphs depicting variation of (i) gravitational potential energy (P.E.), (ii) kinetic energy (K.E.),and (iii) the total sum of potential and kinetic energies for a freely falling body are as shown in adjoining Fig. From the graphs, it is clear that:
(a) Gravitational potential energy decreases as the body falls downwards and is zero at the Earth.

(b) Kinetic energy increases as the body falls downwards and is maximum when the body just strikes the ground.
(c) The sum of kinetic and potential energies remains constant at all points during its free fall.

Let m be the mass of each ball bearing. Before collision, total K.E. of the system 
\(=\frac { 1 }{ 2 } { mv }^{ 2 }+0=\frac { 1 }{ 2 } { mv }^{ 2 }\)
After collision, K.E. of the system is
Case I, \({ E }_{ 1 }=\frac { 1 }{ 2 } (2m){ (v/2) }^{ 2 }=\frac { 1 }{ 4 } { mv }^{ 2 }\)
Case II, \({ E }_{ 2 }=\frac { 1 }{ 2 } { mv }^{ 2 }\)
Case III, \({ E }_{ 3 }=\frac { 1 }{ 2 } (3m)({ v/3) }^{ 2 }=\frac { 1 }{ 6 } { mv }^{ 2 }\)
Thus, case II is the only possibility since K.E. is conserved in this case.

We can calculate spring constant of spring by first information.
  
Work done by a vertical spring is = 0.3 g = k[10\(\times \) 10^-2]
2 = k\(\times \)0.1\(\Rightarrow \) k = \(\frac { 3 }{ 0.1 } =30\frac { N }{ m } \)
Extra work required to stretch it by next 5 cm.
W = \(\frac { 1 }{ 2 } k{ x }_{ 2 }^{ 2 }\)-\(\frac { 1 }{ 2 } k{ x }_{ 1 }^{ 2 }\)
W = \(\frac { 1 }{ 2 } \times 30\left[ { \left( 15\times { 10 }^{ -2 } \right) }^{ 2 }-{ \left( 10\times { 10 }^{ -2 } \right) }^{ 2 } \right] \)
W = 15[225-100] \(\times { 10 }^{ -4 }\) = 15\(\times \)125\(\times { 10 }^{ -4 }\) J
W = 0.1875 J

(a) Power used by family, P = 8 kW = 8000 W
As only 20% of solar energy can be converted to useful electrical energy, hence power to be supplied by solar energy \(\frac { 8000\quad W }{ 20% } =40000\quad W\) As solar energy is incident at a rate of 200 Wm-2 hence the area needed \(A=\frac { 40000\quad W }{ 200\quad W{ m }^{ 2 } } =200{ m }^{ 2 }\) 
(b) The area needed is comparable to roof are of a large sized house.

Work done  by the force = force\( \times \)displacement
or W = F \( \times \) S
But from Newton's 2nd law , we have
Force = mass\( \times \)acceleration
i.e.  F = ma                      ...(ii)
hence , from Eqs (i) and (ii), we get
W = m( \(\frac{d^2s}{dt^2}\) ) S    [ \(\because\) a =  [ \(\frac{d^2s}{dt^2}\)]........(iii)
now, we have
S =  \(\frac{1}{3}\)t²
\(\therefore\)  \(\frac{d^2s}{dt^2}\) =  \(\frac{d}{dt}\)[\(\frac{d}{dt}\)( \(\frac{1}{3}\)t²)]
=  \( \frac{d}{dt}\) \( \times \)( \(\frac{2}{3}\)t )
=  \( \frac{2}{3}\)\(\frac{dt}{dt}\)
=  \(\frac{2}{3}\)
Hence Eq. (iii) becomes
W = \( \frac{2}{3} \)ms
= \(\frac{2}{3}\)\( \times \) m\( \times \)\(\frac{1}{3}\)t²
= \(\frac{2}{9}\)mt²
We have , m = 3 kg , t = 2s
\(\therefore\)      W = \(\frac{2}{9}\)\( \times \)3\( \times \)(2)²
= \(\frac{8}{3}\) J

The total (net) momentum can change only when some net external force acts on the system. Hence, its rate of change is proportional to the external force.

Friction always opposes motion. Hence, work done by a body against friction cause loss in its kinetic energy

The potential energy decrease, because 
\(\Delta\)U=-\(\int { F.dx } \)If work done by conservation force is p[ositive, then obviously \(\Delta\)U is -ve

(i) Area swept by blades of windmill = A and wind velocity = v|
\(\therefore \) Volume of air passing per unit time = Av
\(\therefore \) Mass of air passing per unit time = Avp
and mass of air passing in time t, M = Avpt
(ii) KE of said quantity of air , K = \(\frac { 1 }{ 2 } { Mv }^{ 2 }=\frac { 1 }{ 2 } { Aptv }^{ 3 }\)
(iii) 

If efficiency of windmill be 25%, then Output electrical power = 25% of input power
\(\frac { 25 }{ 100 } \times \frac { 1 }{ 2 } Ap{ v }^{ 3 }\)

As \(A={ 30m }^{ 2 },\ v=36\ km/h\ =36\times \frac { 5 }{ 18 } m/s\)
\(\\ =10m/s\ and\ p=1.2\ { kgm }^{ -3 }\)
\(\\ \therefore Output\ electrical\ power=\frac { 25 }{ 100 } \times \frac { 1 }{ 2 } \times 30\times 1.2\times ({ 10 })^{ 3 }\)
\(\\ =4500\quad W=4.5\ kW\ 1kW=1000W]\)