### Related Formula sum F = ma$\sum F = ma$ a_1 = textConstraint relationtimes a_2$a_1 = \text{Constraint relation}\times a_2$ ### Core Logic

Let the tension in the string attached to the 2 mathrm~kg$2 \mathrm{~kg}$ block be T$T$. By tracing the string around the movable pulley, the tension supporting the 4 mathrm~kg$4 \mathrm{~kg}$ mass becomes 2T$2T$. From the principle of virtual work (or string constraints), if the 4 mathrm~kg$4 \mathrm{~kg}$ block moves down with an acceleration a$a$, the string shortens by 2x$2x$ on the incline side, meaning the 2 mathrm~kg$2 \mathrm{~kg}$ block moves up the incline with an acceleration of 2a$2a$. ### Step 1: Write Equations of Motion For the 4 mathrm~kg$4 \mathrm{~kg}$ block (moving downwards): 4g - 2T = 4a$4g - 2T = 4a$ 40 - 2T = 4a quad dots (i)$40 - 2T = 4a \quad \dots (i)$ For the 2 mathrm~kg$2 \mathrm{~kg}$ block (moving up the incline): T - 2g sin(30^circ) = 2(2a)$T - 2g \sin(30^\circ) = 2(2a)$ T - 2(10)left(frac12right) = 4a$T - 2(10)\left(\frac{1}{2}\right) = 4a$ T - 10 = 4a quad dots (ii)$T - 10 = 4a \quad \dots (ii)$ ### Step 2: Solve for Acceleration From equation (ii), we have 4a = T - 10$4a = T - 10$. Substitute this directly into equation (i) or add the appropriately multiplied equations: 40 - 2T = T - 10$40 - 2T = T - 10$ 3T = 50 Rightarrow T = frac503 mathrm~N$3T = 50 \Rightarrow T = \frac{50}{3} \mathrm{~N}$ Substitute T$T$ back into (ii): frac503 - 10 = 4a$\frac{50}{3} - 10 = 4a$ 4a = frac203 Rightarrow a = frac53 mathrm~m/s^2$4a = \frac{20}{3} \Rightarrow a = \frac{5}{3} \mathrm{~m/s^2}$ We need the acceleration of the 2 mathrm~kg$2 \mathrm{~kg}$ block, which is 2a$2a$: 2a = 2 times left(frac53right) = frac103 mathrm~m/s^2$2a = 2 \times \left(\frac{5}{3}\right) = \frac{10}{3} \mathrm{~m/s^2}$ Since g = 10 mathrm~m/s^2$g = 10 \mathrm{~m/s^2}$, this acceleration is exactly fracg3$\frac{g}{3}$. ### Pattern Recognition Movable pulleys double the force but halve the displacement/acceleration on the supported side. Tension on the movable pulley side is twice the tension on the single string side. Remember to explicitly solve for the requested block's specific acceleration, not just 'a'. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula f_k = mu mg cos theta$f_k = \mu mg \cos \theta$ F_1 = mg sin theta + f_k$F_1 = mg \sin \theta + f_k$ F_2 = mg sin theta - f_k$F_2 = mg \sin \theta - f_k$ ### Core Logic To move the block up, the applied force F_1$F_1$ must overcome both the downward gravitational component and the downward frictional force. To prevent it from sliding down, the applied force F_2$F_2$ acts upwards and is aided by friction which acts upwards to oppose impending downward slip.

### Step 1: Calculate Friction f_k = mu mg cos theta$f_k = \mu mg \cos \theta$ f_k = 0.1 times 5 times 10 times cos(30^circ)$f_k = 0.1 \times 5 \times 10 \times \cos(30^{\circ})$ f_k = 5 times fracsqrt32 = 2.5sqrt3 text N$f_k = 5 \times \frac{\sqrt{3}}{2} = 2.5\sqrt{3} \text{ N}$ ### Step 2: Force Equations Moving up: F_1 = mg sin theta + f_k = 50 sin(30^circ) + 2.5sqrt3 = 25 + 2.5sqrt3$F_1 = mg \sin \theta + f_k = 50 \sin(30^{\circ}) + 2.5\sqrt{3} = 25 + 2.5\sqrt{3}$ Preventing slip down: F_2 = mg sin theta - f_k = 50 sin(30^circ) - 2.5sqrt3 = 25 - 2.5sqrt3$F_2 = mg \sin \theta - f_k = 50 \sin(30^{\circ}) - 2.5\sqrt{3} = 25 - 2.5\sqrt{3}$ ### Step 3: Difference calculation |F_1| - |F_2| = (25 + 2.5sqrt3) - (25 - 2.5sqrt3)$|F_1| - |F_2| = (25 + 2.5\sqrt{3}) - (25 - 2.5\sqrt{3})$ |F_1| - |F_2| = 2 times 2.5sqrt3 = 5sqrt3 text N$|F_1| - |F_2| = 2 \times 2.5\sqrt{3} = 5\sqrt{3} \text{ N}$ *Note: The official options had an anomaly where 5sqrt3 text N$5\sqrt{3} \text{ N}$ was missing or evaluated as a bonus. Option 2 was listed as 50sqrt3$50\sqrt{3}$ in the primary text. We track the closest logic path indicating Bonus.* ### Pattern Recognition The difference between 'push up' and 'hold from sliding' forces on an incline is always precisely 2 f_k$2 f_k$ (2 mu mg cos theta$2 \mu mg \cos \theta$). Bypass calculating the mg sin theta$mg \sin \theta$ terms entirely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula sum F = ma$\sum F = ma$ f_k = mu_k N = mu_k mg costheta$f_k = \mu_k N = \mu_k mg \cos\theta$ ### Core Logic

Since block M$M$ moves down the incline, kinetic friction opposes its motion (acts upwards). Block m$m$ is pulled up the incline, so kinetic friction opposes its motion (acts downwards). For M$M$ block (53^circ$53^\circ$ slope): Mg sin 53^circ - mu Mg cos 53^circ - T = Ma$Mg \sin 53^{\circ} - \mu Mg \cos 53^{\circ} - T = Ma$ ### Step 1: Tension Calculation Given M = 10mathrm\,kg$M = 10\mathrm{\,kg}$, a = 2mathrm\,m/s^2$a = 2\mathrm{\,m/s^2}$, mu = 0.25$\mu = 0.25$, g = 10mathrm\,m/s^2$g = 10\mathrm{\,m/s^2}$. sin 53^circ = 4/5 = 0.8$\sin 53^\circ = 4/5 = 0.8$, cos 53^circ = 3/5 = 0.6$\cos 53^\circ = 3/5 = 0.6$. 10(10)(0.8) - 0.25(10)(10)(0.6) - T = 10(2)$10(10)(0.8) - 0.25(10)(10)(0.6) - T = 10(2)$ 80 - 15 - T = 20$80 - 15 - T = 20$ 65 - T = 20 Rightarrow T = 45mathrm\,N$65 - T = 20 \Rightarrow T = 45\mathrm{\,N}$ ### Step 2: Evaluate mass m For m$m$ block (37^circ$37^\circ$ slope) moving upward: T - mg sin 37^circ - mu mg cos 37^circ = ma$T - mg \sin 37^{\circ} - \mu mg \cos 37^{\circ} = ma$ sin 37^circ = 3/5 = 0.6$\sin 37^\circ = 3/5 = 0.6$, cos 37^circ = 4/5 = 0.8$\cos 37^\circ = 4/5 = 0.8$. 45 - m(10)(0.6) - 0.25(m)(10)(0.8) = m(2)$45 - m(10)(0.6) - 0.25(m)(10)(0.8) = m(2)$ 45 - 6m - 2m = 2m$45 - 6m - 2m = 2m$ 45 - 8m = 2m$45 - 8m = 2m$ 10m = 45 Rightarrow m = 4.5mathrm\,kg$10m = 45 \Rightarrow m = 4.5\mathrm{\,kg}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws Of Motion

### Related Formula f_s leq mu_s N$f_s \leq \mu_s N$ F_c = mromega^2$F_c = mr\omega^2$ ### Core Logic

To prevent the coin from slipping, the static friction must provide the necessary centripetal force for circular motion. f = momega^2 r$f = m\omega^2 r$ The normal force on the flat disc is N = mg$N = mg$. The maximum static friction is f_textmax = mu N = mu mg$f_{\text{max}} = \mu N = \mu mg$. For no slipping: m r omega^2 leq mu mg$m r \omega^2 \leq \mu mg$ omega^2 leq fracmu gr$\omega^2 \leq \frac{\mu g}{r}$ omega_textmax = sqrtfracmu gr$\omega_{\text{max}} = \sqrt{\frac{\mu g}{r}}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws Of Motion