Work, Energy and Power

$\omega =\Delta k.E=\frac{1}{2}m\left[v_t^2-v_i^2\right]$

$=\frac{1}{2}m{\left[b*4^{5/2}\right]}^2-\frac{1}{2}m{\left[0\right]}^2$

$=\frac{1}{2}*0.5\left[0.25^2*4^5\right]=2^4=16J$

$KE=\frac{{\rho }^2}{2m}$

$\Rightarrow {\rho }^2\alpha m$

$\Rightarrow {\left(\frac{P_1}{P_2}\right)}^2=\frac{8}{2}=\frac{4}{1}$

$\Rightarrow \frac{P_1}{P_2}=\frac{2}{1}$

m = 200 g, ki = 90 J kf = 40 J.,  $\Delta t=1sec$

$\Rightarrow F=\frac{k_f-k_i}{x}\Rightarrow \sqrt[]{.4*90}-\sqrt[]{.4*40}=F*1$

$\Rightarrow F=6-4=2N$

So for complete rest :-

Fx = 90 J

2x = 90 x = 45 m

Using conservation of momentum:

60 * v = (60 + 120) * 2

⇒  v = 6m/s

From conservation of Energy:

$mg\mathcal{l}\left(1-cos60°\right)=\frac{1}{2}m{v}^2$

$\Rightarrow v^2=2g\mathcal{l}\left(\frac{1}{2}\right)=g\mathcal{l}$

$=10*\frac{25}{10}=5m/s$

6.30 The decay process of free neutron at rest is given as:

n $?p+\stackrel{?}{e}$

From Einstein's mass-energy relation, we have the energy of electron as?  $m{c}^2$ where,

? m = Mass defect = Mass of neutron – ( Mass of proton + mass of electron)

c = speed of light

? m and c are constant. Hence, the given two-body decay is unable to explain the continuous energy distribution in the $?$ -decay $$ of a neutron or a nucleus. The presence of neutrino von the LHS of the decay correctly explain the continuous energy distribution.

6.29 The potential energy of two masses in a system is inversely proportional to the distance between them. The potential energy of the system of two balls will decrease as they get closer to each other. When the balls touch each other, the potential energy becomes zero, I.e. at r = 2R. The potential energy curve in (i), (ii), (iii), (iv) and (vi) do not satisfy these conditions. So there is no elastic collision.

6.28 Mass, m = 200 kg

Speed, v = 36 km/h = 10 m/s

Mass of the boy, M = 20 kg

Initial momentum = (M + m)v = (20 + 200) x 10 kg-m/s = 2200 kg-m/s

If v' is the final velocity of the trolley, then

The final momentum = (M+m) x v' – M x 4= 220v'-80

According to the law of conservation of energy,

Initial momentum = final momentum

2200 = 220v'-80

V' = 10.36 m/s

Time required by the boy to travel 10m = 10/4 = 2.5 s

Distance travel by trolley in 2.5 s = 10.36 x 2.5 m = 25.9 m

6.27 Mass of the bolt, m = 0.3 kg, Height of the elevator, h = 3 m

Since the bolt did not rebound, the entire potential energy got converted into heat.

The potential energy of the bolt = mgh = 0.3 x 9.8 x 3 J = 8.82 J

The heat produced will remain same even if the lift is stationary, since g = constant

6.26 Mass of the block = 1 kg

Spring constant = 100 N/m

Displacement of the block, x = 10 cm = 0.1 m

At equilibrium, normal reaction, R = $mg\cos ?37°$

Frictional force, F = $?R$ = $mg\sin ?37°$

Net force acting on the block down on the incline = $mg\sin ?37°$ - F

= $mg\sin ?37°$ - $?mg\cos ?37°$

=mg ( $\sin ?37°-?\cos ?37°$ )

At equilibrium,

Work done = Potential energy of the stretched string

mg ( $\sin ?37°-?\cos ?37°$ ) = (1/2)kx2

1 x 10 x ( $\sin ?37°-?\cos ?37°$ ) = (1/2) x 100 x 0.1

10 x (0.602 – $?0.798)$ = 0.5 x 100 x 0.1

$?=0.128$