$\frac{1}{2}m(2)^{2} = F \times x \implies K = F~x$
$2K = F~x^{\prime} \implies x^{\prime} = 2x$
$10 \times 1 = (0.2)(10)x$
$x = \frac{10}{2} = 5~m$
Detailed solution not required.
Observing force is in radial direction. Hence its work is zero.
At equilibrium: $kx = 3mg$, $T = mg$
When string is cut $T = 0$.
For 2m: $3~mg - 2~mg = 2~ma \implies a = \frac{g}{2}$ upward.
For m: $mg = ma \implies a = g$ downward.
$\frac{1}{2}mv_{0}^{2} = mgL~\sin~\theta + \mu mgL~\cos~\theta$
$\frac{1}{2}m\frac{v_{0}^{2}}{4} = mgL~\sin~\theta - \mu mgL~\cos~\theta$
$\frac{4}{1} = \frac{\sin~\theta + \mu~\cos~\theta}{\sin~\theta - \mu~\cos~\theta}$
$5\mu = \frac{3}{\sqrt{3}} \implies \mu = \frac{\sqrt{3}}{5} = \frac{1}{1000} \times I$
$I = (1.732)\frac{1000}{5} \approx 346.4$
$kx = 10 \times 12 = 120~N$
For 20 kg mass:
$200 - 120 = 20a \implies a = 4~m/s^{2}$
Detailed solution not required.
$N_{A}\sin~60^{\circ} = 50g \implies N_{A} = \frac{1000}{\sqrt{3}}N$
$N_{B} = N_{A}\cos~60^{\circ} = \frac{500}{\sqrt{3}}N$
Let tension in the string = T.
$T = \frac{F}{2}$ (pulley is massless)
$(T - f) = ma \implies a = \left(\frac{F}{2} - f\right) / m$
$F_{up} = mg~\sin~\phi + \mu~mg~\cos~\phi$
$F_{Down} = mg~\sin~\phi - \mu~mg~\cos~\phi$
$2F_{Down} = F_{up}$
$2~mg~\sin~\phi - 2\mu~mg\cos~\phi = mg~\sin~\phi + \mu~mg~\cos~\phi$
$mg~\sin~\phi = 3\mu~mg~\cos~\phi \implies \tan~\phi = 3\mu = 3~\tan~\theta$ (since $\mu = \tan~\theta$)
$T = \frac{F}{2} = \frac{60}{2} = 30~N$
For 2 kg block: $30 - 2 \times 10 = 2a_{1} \implies a_{1} = 5~m/s^{2}$ (upward)
For 4 kg block: $30 - 4 \times 10 = 4a_{2} \implies a_{2} = -2.5~m/s^{2}$ (downward)
Speed at lowest point $v_{0} = \sqrt{2gL}$
For completing the circle about N:
$v_{0} \ge \sqrt{5g(L - h)}$
$2gL \ge 5g(L - h) \implies 5gh \ge 3gL \implies h \ge \frac{3}{5}L$
For downward motion: $mg~\sin~\theta = \mu~mg~\cos~\theta$
Retardation in upward motion $= \frac{mg~\sin~\theta + \mu~mg\cos~\theta}{m} = 2g~\sin~\theta$
Distance covered $= \frac{v_{0}^{2}}{2(2g~\sin~\theta)} = \frac{v_{0}^{2}}{4g~\sin~\theta}$
Detailed solution not required.
$T = \frac{3}{2}mg$
For mass 'm': $\frac{3}{2}mg - mg = ma \implies a = \frac{g}{2}$
With respect to pulley: $2mg + 2ma - T = 2m\left(\frac{g}{2}\right)$
$T - mg - ma = m\frac{g}{2} \implies mg + ma = 3m\frac{g}{2} \implies a = \frac{g}{2}$
$T - 60g - T_{1} = 60 \times 2$ ... (1)
$T_{1} = T_{2} + 50g$ ... (2)
$40g - T_{2} = 40 \times 1$ ... (3)
$T_{2} = 400 - 40 = 360$
$T_{1} = 360 + 500 = 860$
$T = 600 + 860 + 120 = 1580~N$
Tensions in corresponding strings are equal to forces applied by the persons.
Horizontal forces must be balanced:
$F_{1}\sin~60^{\circ} = F_{2}\cos~60^{\circ} + F_{3} \implies \sqrt{3}F_{1} = F_{2} + 2F_{3}$
Detailed solution not required.
Detailed solution not required.
$f_{mg} = (1 - f)mg \times \mu$
$f = \frac{(1 - f)}{4} \implies 4f = 1 - f \implies 5f = 1 \implies f = \frac{1}{5}$
In $\% = \frac{1}{5} \times 100 = 20\%$
Detailed solution not required.
Detailed solution not required.
$\vec{v} = \vec{u} + \frac{\vec{F}}{m}t$
$\vec{v} = 6\hat{i} - 12\hat{j} + \frac{-3\hat{i} + 4\hat{j}}{5}t$
If body has velocity just along y-axis, the coefficient of $\hat{i}$ will be zero.
$6 - \frac{3}{5}t = 0 \implies t = \frac{6 \times 5}{3} = 10~s$
$|\vec{F}| = \sqrt{1^{2} + 1^{2} + (\sqrt{2})^{2}} = 2~N$
$\cos~\gamma = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \implies \gamma = 45^{\circ}$
For 200 g block:
$T - 0.15 \times 0.2 \times 9.8 = 0.2 \times 0.30 \implies T = 0.294 + 0.06 = 0.354~N$
For 700 g block:
$F - T - 0.15 \times 0.2 \times 9.8 - 0.15 \times 0.9 \times 9.8 = 0.7 \times 0.30$
$F = 0.354 + 0.294 + 1.323 = 2.18~N$
$\vec{r} = 3t\hat{i} - 4~\cos~t\hat{j}$
$\vec{v} = 3\hat{i} + 4~\sin~t\hat{j}$
$\vec{a} = 4~\cos~t\hat{j}$
$\vec{F} = m\vec{a} = 12~\cos~t\hat{j}$
Impulse $= \int_{0}^{\pi/2} 12~\cos~t\hat{j}~dt = 12[(\sin~t)]_{0}^{\pi/2}\hat{j} = 12\hat{j}~N\cdot s$
Let acceleration of cart be a.
$N = ma$
$\mu~ma = mg \implies a = \frac{g}{\mu}$
Detailed solution not required.
$x = A~\sin~4\pi t$
$v = 4\pi A~\cos~4\pi t$
$a = -16\pi^{2}A~\sin~4\pi t$
At $t = \frac{1}{8}$:
$a = -16\pi^{2}A~\sin\left(\frac{4\pi}{8}\right) = -16\pi^{2}A~\sin\left(\frac{\pi}{2}\right) = -16\pi^{2}A$
$F = -16\pi^{2}Am$
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
$0.01 \times 500 = 2v_{1} + 0.01v_{2}$ ... (i)
$\frac{1}{2}(2)v_{1}^{2} = 2 \times 10 \times 0.1 \implies v_{1} = \sqrt{2}$
From equation (i):
$5 = 2\sqrt{2} + 0.01v_{2} \implies v_{2} \approx 220~m/s$
$f = mg~\sin~\theta$
$s = vt$ (upward)
$W_{f} = (mg~\sin~\theta)(vt)\cos(90^{\circ} - \theta) = mgvt~\sin^{2}\theta$
At horizontal floor:
$v_{A} = \sqrt{2(4h)g}$, $v_{B} = \sqrt{2gh}$
After elastic collision velocities will be interchanged.
$v_{A} = \sqrt{2gh}$, $v_{B} = \sqrt{8hg}$
$h_{A} = \frac{1}{2}\frac{m \times 2hg}{mg} = h$, $h_{B} = 4h$
But at the end, the particle will leave the wedge:
$\frac{1}{2}mu^{2} = \frac{1}{2}m(8gh) - mgh = 3gh \implies u^{2} = 6gh$
Now $h_{max} = \frac{6gh~\sin^{2}60^{\circ}}{2g} = \frac{6h}{2} \times \frac{3}{4} = \frac{9}{4}h$
Total $h + \frac{9}{4}h = \frac{13h}{4}$
Ratio $h : \frac{13h}{4} = 4 : 13$
Power = energy given per second
$= 3 \times \frac{1}{2} \times 0.05 \times 10^{2} = \frac{15}{2} = 7.5~W$
Detailed solution not required.
$\mu_{b/cB} = v_{0}$
$\mu_{b/cB} = \mu g$ (acceleration)
To reach B: $0^{2} = (v_{0})^{2} - 2\mu gL \implies v_{0} = \sqrt{2\mu gL}$
From conservation of momentum:
$1v_{1} + 2v_{2} = 0 \implies v_{1} = -2v_{2}$
$\frac{1}{2} \times 1(-2v_{2})^{2} + \frac{1}{2} \times 2 \times v_{2}^{2} = 12$
$2v_{2}^{2} + v_{2}^{2} = 12 \implies v_{2} = 2~m/s$
$\frac{10 \times 10^{3} \times 10 \times 20}{20 \times 60 \times P} = \frac{60}{100}$
$P = \frac{10}{3600} = \frac{100}{36} \times 10^{3} = 2.74~kW$
Detailed solution not required.
$a_{T} = g~\sin~\theta$
$= 9.8 \times \frac{\sqrt{3}}{2} = 4.9\sqrt{3}~m/s^{2}$
$v^{2} = 7^{2} - 2 \times 9.8 \times 1 = 49 - 19.6 \approx 29$
$h = \frac{29}{20} = 1.45~m$
Total height $= 1.45 + 1 = 2.45~m$
$m\sqrt{2gh} = 2mv \implies v = \sqrt{\frac{gh}{2}}$
$2mgh^{\prime} = \frac{1}{2}(2m)\left(\sqrt{\frac{gh}{2}}\right)^{2} \implies h^{\prime} = \frac{h}{4}$