Physics 3 Marks Question

$\text{mg}=10\text{N},\mu=0.2,\text{H}=1\text{m},\text{u}=\text{v}=0$

change in P.E. = work done.

Increase in K.E.

⇒ w = mgh = 10 × 1 = 10J

Again, on the horizontal surface the fictional force

$\text{F}=\mu\text{R}=\mu\text{mg}$

$=0.2\times10=2\text{N}$

So, the K.E. is used to overcome friction

$\Rightarrow\text{S}=\frac{\text{W}}{\text{F}}=\frac{10\text{J}}{2\text{N}}=5\text{m}$

1. Initial kinetic energy of the ball, $\text{K}_\text{i}=\frac{1}{2}\text{mv}^2$

Here, m is the mass of the ball.

The final kinetic of the ball is zero.

Work done by the kinetic friction is equal to the change in kinetic energy of the ball.

2. $\therefore$ Work done by the kinetic friction $=\text{K}_\text{f}-\text{K}_\text{i}$

$=0-\frac{1}{2}\text{mv}^2$

$=-\frac{1}{2}\text{mv}^2$

The minimum velocity required to cross the height point $\text{c}=\sqrt{2\text{gl}}$

Let the rod released from a height h.

Total energy at A = total energy at B

$\text{mgh}=\frac{1}{2}\text{mv}^2$

$\text{mgh}=\frac{1}{2}\text{m}(2\text{gl})$

[Because v = required velocity at B such that the block makes a complete circle.]

So, h = l

m = 250g = 0.250kg,

k = 100N/ m, m = 10cm = 0.1m

g = 10m/ sec²

Applying law of conservation of energy,

$=\frac{1}{2}\text{kx}^2=\text{mgh}$

$\Rightarrow\text{h}=\frac{1}{2}\Big(\frac{\text{kx}^2}{\text{mg}}\Big)$

$=\frac{100\times(0.1)^2}{2\times0.25\times10}$

$=0.2\text{m}=20\text{cm}$

m = 100g = 0.1kg, x = 5cm = 0.05m, k = 100N/m

When the body leaves the spring, let the velocity be v,

$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{k}\text{x}^2$

$\Rightarrow\text{v}=\text{x}\sqrt{\frac{\text{k}}{\text{m}}}$

$\Rightarrow0.05\times\sqrt{\frac{100}{0.1}}=1.58\text{m}/\text{sec}$

For the projectile motion, $\theta=0^\circ,\text{Y}=-2$

Now, $\text{Y}=(\text{u}\sin\theta)\text{t}-\frac{1}{2}\text{gt}^2$

$\Rightarrow-2=\Big(\frac{-1}{2}\Big)\times9.8\times\text{t}^2$

$\Rightarrow\text{t}=0.63\text{sec}$

So, $\text{x}=(\text{u}\cos\theta)\text{t}$

$\Rightarrow1.58\times0.63=1\text{m}$

$\text{a}=\frac{\text{v}^2-\text{u}^2}{2\text{s}}$

$=\frac{0.16}{0.4}=0.04\text{m}/\text{sec}^2$

F = 2.50N, S = 2.5m, m = 15g = 0.015kg.

So, $\text{w} =\text{F} \times \text{S}$

$\Rightarrow\text{a}=\frac{\text{F}}{\text{m}}=\frac{2.5}{0.015}$

$=\frac{500}{3}\text{m/s}^2$

$=\text{F}\times\text{S}\cos0^\circ$ (acting along the same line)

$=2.5\times2.5=6.25\text{J}$

Let the velocity of the body at b = U. Applying work-energy principle $\frac{1}{2}\text{mv}^2-0=6.25$

$\Rightarrow\text{V}=\sqrt{\frac{6.25\times2}{0.015}}=28.86\text{m/sec}$

So, time taken to travel from A to B.

$\Rightarrow\text{t}=\frac{\text{v-u}}{\text{a}}=\frac{28.86\times3}{500}$

$\therefore$ Average power $=\frac{\text{W}}{\text{t}}=\frac{6.25\times500}{(28.86)\times3}=36.1$

Given, m = 100g = 0.1kg, v = 5m/sec, r = 10cm

Work done by the block = total energy at A - total energy at B

$\Big(\frac{1}{2}\text{mv}^2+\text{mgh}\Big)-0$

$\Rightarrow\text{W}=\frac{1}{2}\text{mv}^2+\text{mgh}-0$

$=\frac{1}{2}\times(0.1)\times25+(0.1)\times10\times(0.2)$[h = 2r = 0.2m]

$\Rightarrow\text{W}=1.25+0.2​​$

$\Rightarrow\text{W}=1.45\text{J}$

So, the work done by the tube on the body is

W_t = -1.45J

$\theta=37^\circ,$ l = h = natural length

Let the velocity when the spring is vertical be ‘v’.

$\cos=37^\circ=\frac{\text{BC}}{\text{AC}}=0.8=\frac{4}{5}$

$\text{AC}=(\text{h}+\text{x})=\frac{5\text{h}}{4}$ (because BC = h)

So, $\text{x}=\frac{5\text{h}}{4}-\text{h}=\frac{\text{h}}{4}$

Applying work energy principle $=\frac{1}{2}\text{kx}^2+\frac{1}{2}\text{mv}^2$

$\Rightarrow\text{v}=\text{x}\sqrt{\Big(\frac{\text{k}}{\text{m}}\Big)}=\frac{\text{h}}{4}\sqrt{\frac{\text{k}}{\text{m}}}$

Given, N = mg

As shown in the figure, $\frac{\text{mv}^2}{\text{R}}=\text{mg}$

$\Rightarrow\text{v}^2=\text{gR}\ \dots(1)$

Total energy at point A = energy at P

$\frac{1}{2}\text{kx}^2=\frac{\text{mgR}+2\text{mgR}}{2}$ [because v² = gR]

$\Rightarrow\text{x}^2=\frac{{3\text{mgR}}}{\text{k}}$

$\Rightarrow\text{x}=\sqrt{\frac{{3\text{mgR}}}{\text{k}}}$

Applying law of conservation of Energy for point A & B

$\text{mgH}=\frac{1}{2}\text{mv}^2+\text{mgh}$

$\Rightarrow\text{g}=\frac{1}{2}\text{v}^2+0.5\text{g}$

$\Rightarrow\text{v}^22(\text{g}-0.59)=\text{g}$

$\Rightarrow\text{v}=\sqrt{\text{g}}=3.1\text{m}/\text{s}$

After point B the body exhibits projectile motion for which

$\theta=0^\circ,\text{v}=-0.5$

So, $-0.5=(\text{u}\sin\theta)\text{t}-\Big(\frac{1}{2}\Big)\text{gt}^2$

$\Rightarrow0.5=4.9\ \text{t}^2$

$\Rightarrow\text{t}=0.31\text{sec}$

So, $\text{x}=(\text{v}\cos\theta)\text{t}$

$=3.1\times3.1=1\text{m}$

So, the particle will hit the ground at a horizontal distance in from B.

m = 2kg, s₁ = 4.8m, R = 20cm = 0.2m, s₂ = 1m,

$\sin37^\circ=0.60=\frac{3}{5},\theta=37^\circ,$

$ \cos37^\circ=.79=0.8=\frac{4}{5},\text{g}=10\text{m}/\text{sec}^2$

Applying work – Energy principle for downward motion of the body

$0-0=\text{mg}\sin37^\circ\times5-\mu\text{R}\times5-\frac{1}{2}\text{kx}^2$

$\Rightarrow20\times(0.60)\times1-\mu\times20\times(0.80)\times1+\frac{1}{2}\text{k}(0.2)^2=0$

$\Rightarrow60-80\mu-0.02\text{k}=0​​$

$\Rightarrow80\mu+0.02\text{k}=60\ \dots(1)$

Similarly, for the upward motion of the body the equation is,

$0-0=(-\text{mg}\sin37^\circ)\times1-\mu\text{R}\times1+\frac{1}{2}\text{k}(0.2)^2$

$\Rightarrow-20\times(0.60)\times1-\mu\times20\times(0.80)\times1+\frac{1}{2}\text{k}(0.2)^2=0$

$\Rightarrow-12-16\mu+0.02\text{K}=0\ \dots(2)$

Adding equation (i) & equation (ii), we get $96\mu=48$

$\Rightarrow\mu=0.5$

Now putting the value of $\mu$ in equation (1), K = 1000N/m

Let ‘dx’ be the length of an element at a distance × from the table

mass of ‘dx’ length $=\Big(\frac{\text{m}}{\ell}\Big)\text{dx}$

Work done to put dx part back on the table,

$\text{W}=\Big(\frac{\text{m}}{\ell}\Big)\text{dx }\text{g(x)}$

So, total work done to put $\frac{\ell}{3}$ part back on the table,

$\text{W}=\int\limits_0^{\frac{1}{3}}\Big(\frac{\text{m}}{\ell}\Big)\text{gx dx}$

$\Rightarrow\text{W}=\Big(\frac{\text{m}}{\ell}\Big)\text{g}\Big[\frac{\text{x}^2}{2}\Big]_0^\frac{\ell}{3}$

$\Rightarrow\frac{\text{mg}\ell^2}{18\ell}=\frac{\text{mg}\ell}{18}$