### Related Formula N + Fsintheta = mg implies N = mg - Fsintheta$N + F\sin\theta = mg \implies N = mg - F\sin\theta$ Fcostheta = f_k = mu_k N$F\cos\theta = f_k = \mu_k N$ W = F cdot S cdot costheta$W = F \cdot S \cdot \cos\theta$ ### Core Logic Since the block moves with uniform velocity, horizontal acceleration is zero. Fcos(45^circ) = mu [mg - Fsin(45^circ)]$F\cos(45^\circ) = \mu [mg - F\sin(45^\circ)]$ fracFsqrt2 = 0.25 left[25 times 9.8 - fracFsqrt2right]$\frac{F}{\sqrt{2}} = 0.25 \left[25 \times 9.8 - \frac{F}{\sqrt{2}}\right]$ fracFsqrt2 = 61.25 - 0.25 fracFsqrt2 implies 1.25 fracFsqrt2 = 61.25$\frac{F}{\sqrt{2}} = 61.25 - 0.25 \frac{F}{\sqrt{2}} \implies 1.25 \frac{F}{\sqrt{2}} = 61.25$ ### Step 1: Compute Force and Work Done fracFsqrt2 = frac61.251.25 = 49 implies F = 49sqrt2text N$\frac{F}{\sqrt{2}} = \frac{61.25}{1.25} = 49 \implies F = 49\sqrt{2}\text{ N}$ The work done by the external force over displacement S=5text m$S=5\text{ m}$ is: W = F S cos(45^circ) = (49sqrt2) times 5 times frac1sqrt2 = 245text J$W = F S \cos(45^\circ) = (49\sqrt{2}) \times 5 \times \frac{1}{\sqrt{2}} = 245\text{ J}$

### Pattern Recognition When uniform velocity is sustained, work done by the external pulling component perfectly matches the work consumed against internal friction dissipation (W = f_k cdot S$W = f_k \cdot S$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula Horizontal equilibrium equation: T_1 costheta_1 = T_2 costheta_2$T_1 \cos\theta_1 = T_2 \cos\theta_2$ Vertical balancing equation: T_1 sintheta_1 + T_2 sintheta_2 = mg$T_1 \sin\theta_1 + T_2 \sin\theta_2 = mg$ ### Core Logic Substitute T_1 = sqrt3T_2$T_1 = \sqrt{3}T_2$ into horizontal equilibrium: sqrt3T_2 costheta_1 = T_2 costheta_2 implies sqrt3costheta_1 = costheta_2$\sqrt{3}T_2 \cos\theta_1 = T_2 \cos\theta_2 implies \sqrt{3}\cos\theta_1 = \cos\theta_2$ ### Step 1: Audit Options Pattern Match Let's check option (2) where theta_1 = 60^circ$\theta_1 = 60^{\circ}$ and theta_2 = 30^circ$\theta_2 = 30^{\circ}$: sqrt3cos(60^circ) = sqrt3 cdot frac12 = fracsqrt32 $\sqrt{3}\cos(60^{\circ}) = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} $ cos(30^circ) = fracsqrt32 $ cos(30^{\circ}) = \frac{\sqrt{3}}{2} $ This perfectly matches the horizontal check expression constraints. ### Step 2: Solve for T2 Now map the variables into the vertical component balancing equation: T_2 left[ sqrt3sin(60^circ) + sin(30^circ) ight] = mg$ T_2 \left[ \sqrt{3}\sin(60^{\circ}) + \sin(30^{\circ}) ight] = mg$ T_2 left[ sqrt3left(fracsqrt32 ight) + frac12 ight] = mg$T_2 \left[ \sqrt{3}\left(\frac{\sqrt{3}}{2} ight) + \frac{1}{2} ight] = mg$ T_2 left[ frac32 + frac12 ight] = mg implies T_2 (2) = mg implies T_2 = fracmg2$T_2 \left[ \frac{3}{2} + \frac{1}{2} ight] = mg implies T_2 (2) = mg implies T_2 = \frac{mg}{2}$ ### Pattern Recognition Lami's Theorem or standard rectangular splitting. When tension is scaled by sqrt3$\sqrt{3}$, it directly hints at complementary 30^circ-60^circ$30^{\circ}-60^{\circ}$ geometric alignments. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula The maximum velocity limit preventing outer slide breakout on a rough banked plane profile is given by standard centrifugal force equations: v_0 = sqrtrg left(fractantheta + mu1 - mutantheta ight)$v_{0} = \sqrt{rg \left(\frac{\tan\theta + \mu}{1 - \mu\tan\theta} ight)}$ ### Core Logic By writing out the components shown in the free body layout

: Nsintheta + fcostheta = fracmv_0^2r$N\sin\theta + f\cos\theta = \frac{mv_{0}^{2}}{r}$ Ncostheta - fsintheta = mg$N\cos\theta - f\sin\theta = mg$ ### Step 1: Isolate Coefficient of Friction Squaring the velocity boundary relation gives : fracv_0^2rg = fractantheta + mu1 - mutantheta $\frac{v_{0}^{2}}{rg} = \frac{\tan\theta + \mu}{1 - \mu\tan\theta} $ Cross multiply to isolate the variable terms : v_0^2 - mu v_0^2tantheta = rgtantheta + mu rg$v_{0}^{2} - \mu v_{0}^{2}\tan\theta = rg\tan\theta + \mu rg$ v_0^2 - rgtantheta = mu(rg + v_0^2tantheta)$v_{0}^{2} - rg\tan\theta = \mu(rg + v_{0}^{2}\tan\theta)$ mu = fracv_0^2 - rgtanthetarg + v_0^2tantheta$\mu = \frac{v_{0}^{2} - rg\tan\theta}{rg + v_{0}^{2}\tan\theta}$ ### Pattern Recognition Maximum limits always balance with plus signs in the numerator fraction. Simply rearrange that template expression directly to map the targeted parameter. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula By Newton's second law of motion, the net upward force acting on an accelerating balloon system is given by: F_textbuoyant - m_texttotal g = m_texttotal a$F_{\text{buoyant}} - m_{\text{total}} g = m_{\text{total}} a$ ### Core Logic Let F$F$ be the constant buoyant force acting upward on the balloon. **Case 1 (Initial upward acceleration)** : F - M g = M a implies F = M(g + a) quad text$F - M g = M a \implies F = M(g + a) \quad \text{}$ **Case 2 (After releasing mass x$x$)** : The new total mass becomes (M - x)$(M - x)$, and its acceleration increases to 3a$3a$: F - (M - x)g = (M - x)3a quad text$F - (M - x)g = (M - x)3a \quad \text{}$ Substitute the value of F$F$ from Case 1 into Case 2 [cite: 836, 839]: M(g + a) - (M - x)g = (M - x)3a$M(g + a) - (M - x)g = (M - x)3a$ M g + M a - M g + x g = 3 M a - 3 x a quad text$M g + M a - M g + x g = 3 M a - 3 x a \quad \text{}$ M a + x g = 3 M a - 3 x a$M a + x g = 3 M a - 3 x a$ x(g + 3a) = 2 M a$x(g + 3a) = 2 M a$ x = frac2 M a3a + g quad text$x = \frac{2 M a}{3a + g} \quad \text{}$ ### Step 1: Visual Context The free-body force layout for both accelerating phases is shown below:

### Pattern Recognition Since the upward buoyant force is completely determined by the balloon's volume, it remains constant. Expressing this constant force in terms of the initial conditions allows you to quickly solve for mass changes when acceleration states vary. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion