Physics 2 Marks Questions

m = 500kg, s = 25m, u = 72km/ h = 20 m/ s,

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

$\Rightarrow\frac{400}{50}=8\text{m}/\text{sec}^2$

Frictional force F = ma = 500 × 8 = 4000N

Let the velocity of the body at A is ‘V’ for minimum velocity given at A velocity of the body at point B is zero. 

Applying law of conservation of energy at A & B.

$\frac{1}{2}\text{mv}^2=\text{mg}(2\ell)$

$\Rightarrow\text{v}=\sqrt{(4\text{g}\ell)}=2\sqrt{\text{g}\ell}$

M_b = 2kg

u = 10m/s

a = 3m/s²

t = 5s

ν = u + at 

= 10 + 3 × 5 = 25m/s

$\therefore$ Final $\text{K.E.}=\frac{1}{2}\text{mv}^2$

$=\frac{1}{2}\times2\times625=625\text{J}$

Let the velocity of the body at A be v

So, the velocity of the body at B is $\frac{\text{v}}{2}$

Energy at point A = Energy at point B

So, $\frac{1}{2}\text{mv}_\text{A}^2=\frac{1}{2}\text{mv}_\text{B}^2-\frac{1}{2}\text{kx}^{2+}$

$\Rightarrow\frac{1}{2}\text{kx}^2=\frac{1}{2}\text{mv}_\text{A}^2-\frac{1}{2}\text{mv}_\text{B}^2$

$\Rightarrow\text{kx}^2=\text{m}\Big(\text{v}_\text{A}^{2+-}\text{v}_\text{B}^2\Big)$

$\Rightarrow\text{kx}^2=\text{m}\Big(\text{v}^2-\frac{\text{v}^2}{4}\Big)$

$\Rightarrow\text{k}=\frac{3\text{m}\text{v}^2}{4\text{x}^2}$

m = 1400kg, v = 54km/h = 15m/sec, h = 10m

Work done = Total K.E. - Total P.E.

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

$=\frac{1}{2}\times1400\times(15)^2-1400\times9.8\times10$

$=157500-137200$

$=20300$

So, work done against friction, W_t = 20300J

$\vec{\text{r}}_1=2\hat{\text{i}}+3\hat{\text{j}}$

$\vec{\text{r}}_2=3\hat{\text{i}}+2\hat{\text{j}}$

$\vec{\text{r}}=\vec{\text{r}}_1-\vec{\text{r}}_2$

$\Rightarrow\vec{\text{r}}=\Big(3\hat{\text{i}}+2\hat{\text{j}}\Big)-\Big(2\hat{\text{i}}+3\hat{\text{j}}\Big)$

$=\hat{\text{i}}-\hat{\text{j}}$

Let the compression be x.

According to law of conservation of energy,

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

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

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

No. It will be in the opposite direction and magnitude will be less due to loss in spring.

m_b = 2kg, s = 40m, a = 0.5m/sec²

So, force applied by the man on the box

F = m_ba = 2 × (0.5) = 1N

$\omega=\text{FS}=1\times40=40\text{J}$

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

$\Rightarrow10\times40+\frac{1}{2}\times2500=\frac{1}{2}\text{v}^2$

$\Rightarrow\text{v}^2=3300$

$\Rightarrow\text{v}=57.4\text{m}/\text{sec}\approx58\text{m}/\text{sec}$

$\text{p}=\frac{\text{w}}{\text{t}},$ Work W = pt = 460 × 117.56J

$=177.6\ \text{J}\approx118\text{J}$

$\text{W}=\int\limits_0^\text{d}\text{Fdx}$

$\text{W}=\int\limits_0^\text{d}\text{(a+bx)dx}$

$=\Big[\text{ax}+\Big(\frac{\text{bx}^2}{2}\Big)\Big]_0^\text{d}$

$=\Big[\text{a}+\frac{1}{2}\text{bd}\Big]\text{d}$

$\text{m} = 5\text{kg}$

$\theta=30^\circ$ 

$\text{S} = 10\text{m}$

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

So, work done by the force of gravity

$\omega=\text{mgh}$

$= 5\times9.8\times5 = 245\text{J}$

The body is displaced x towards right.

Let the velocity of the body be v at its mean position.

Applying law of conservation of energy.

$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{k}_1\text{x}^2+\frac{1}{2}\text{k}_2\text{x}^2$

$\Rightarrow\text{mv}^2=\text{x}^2(\text{k}_1+\text{k}_2)$

$\Rightarrow\text{v}^2=\frac{\text{x}^2(\text{k}_1+\text{k}_2)}{\text{m}}$

$\Rightarrow\text{v}=\text{x}\sqrt{\frac{\text{k}_1+\text{k}_2}{\text{m}}}$

m = 2000kg, s = 12m, t = 1min = 60sec

So, work $\text{W}=\text{F}\cos\theta=\text{mgs}\cos0^\circ$ $[\theta=0^\circ,$ for minimum work$]$

$=2000\times10\times12=240000\text{J}$

So, power $\text{p}=\frac{\text{W}}{\text{t}}=\frac{240000}{60}=4000\text{watt}$

h.p $=\frac{4000}{746}=5.3\text{hp}$

m = 500kg, u = 0, v = 72km/h = 20m/s

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

$\Rightarrow\frac{400}{50}=8\text{m}/\text{sec}^2$

force needed to accelerate the car F = ma = 500 × 8 = 4000N

$\text{m}_\text{b}=250\text{g}=.250\text{kg}$

$\theta=37^\circ,\text{S}=1\text{m}.$

Frictional force $\text{f}=\mu\text{R}$

$\text{mg}\sin\theta=\mu\text{R}\ \dots(1)$

$\text{mg}\cos\theta \dots(2)$

so, work done against $\mu\text{R}=\mu\text{RS}\cos0^\circ=\text{mg}\sin\theta$

$\text{S}=0.250\times9.8\times0.60\times1=1.5\text{J}$

M = m_c + m_b = 90kg

u = 6.0km/h = 1.666 m/sec

ν = 12 km/h = 3.333 m/sec

Increase in K.E.

$=\frac{1}{2}\text{Mv}^2-\frac{1}{2}\text{mu}^2$

$=\frac{1}{2}90\times(3.333)^2-\frac{1}{2}90\times(1.66)^2$

$=499.4-124.6$

$=374.8\approx375\text{J}$