A body of mass 1mathrmkg$1\mathrm{kg}$ is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions mathbfT_1$\mathbf{T}_1$ and mathbfT_2$\mathbf{T}_2$ , respectively, are (in N):

### Related Formula veca = fracvecFm$$\vec{a} = \frac{\vec{F}}{m}$$ vecv = int veca \, dt$$\vec{v} = \int \vec{a} \, dt$$ vecr = int vecv \, dt$$\vec{r} = \int \vec{v} \, dt$$ ### Core Logic Given vecF = 4t^3hati - 3thatj$\vec{F} = 4t^3\hat{i} - 3t\hat{j}$ and mass m = 4text kg$m = 4\text{ kg}$. Acceleration: veca = fracvecFm = frac4t^3hati - 3thatj4 = t^3hati - frac34thatj$$\vec{a} = \frac{\vec{F}}{m} = \frac{4t^3\hat{i} - 3t\hat{j}}{4} = t^3\hat{i} - \frac{3}{4}t\hat{j}$$ ### Step 1: Calculating Velocity Velocity is the integral of acceleration with v(0) = 0$v(0) = 0$: vecv = int left( t^3hati - frac34thatj right) dt = fract^44hati - frac38t^2hatj$$\vec{v} = \int \left( t^3\hat{i} - \frac{3}{4}t\hat{j} \right) dt = \frac{t^4}{4}\hat{i} - \frac{3}{8}t^2\hat{j}$$ At t = 2text s$t = 2\text{ s}$: vecv(2) = frac(2)^44hati - frac38(2)^2hatj = 4hati - frac32hatj$$\vec{v}(2) = \frac{(2)^4}{4}\hat{i} - \frac{3}{8}(2)^2\hat{j} = 4\hat{i} - \frac{3}{2}\hat{j}$$ ### Step 2: Calculating Position Position is the integral of velocity with vecr(0) = vec0$\vec{r}(0) = \vec{0}$: vecr = int vecv \, dt = int left( fract^44hati - frac38t^2hatj right) dt$$\vec{r} = \int \vec{v} \, dt = \int \left( \frac{t^4}{4}\hat{i} - \frac{3}{8}t^2\hat{j} \right) dt$$ vecr = fract^520hati - frac324t^3hatj = fract^520hati - fract^38hatj$$\vec{r} = \frac{t^5}{20}\hat{i} - \frac{3}{24}t^3\hat{j} = \frac{t^5}{20}\hat{i} - \frac{t^3}{8}\hat{j}$$ At t = 2text s$t = 2\text{ s}$: vecr(2) = frac(2)^520hati - frac(2)^38hatj = frac3220hati - frac88hatj = frac85hati - hatj$$\vec{r}(2) = \frac{(2)^5}{20}\hat{i} - \frac{(2)^3}{8}\hat{j} = \frac{32}{20}\hat{i} - \frac{8}{8}\hat{j} = \frac{8}{5}\hat{i} - \hat{j}$$ ### Pattern Recognition Recognize F to a to v to r$F \to a \to v \to r$ means consecutive integrations. Because the initial state is from rest at origin, we don't have to worry about constants of integration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

### Related Formula N = Momega^2 R$N = M\omega^2 R$ f_textmax = mu N$$f_{\text{max}} = \mu N$$ f = Mg$f = Mg$ ### Core Logic

For the mass M$M$ to remain stuck to the inner wall and not slide down, the upward frictional force must balance the downward gravitational force. f = Mg$f = Mg$ The normal force N$N$ providing the friction is generated by the centrifugal effect (or provides the centripetal acceleration): N = Momega^2 R$N = M\omega^2 R$ The condition for no sliding is that required friction cannot exceed the maximum static friction: Mg le mu N$Mg \le \mu N$ ### Step 1: Final Conclusion Substitute N$N$ into the inequality: Mg le mu (Momega^2 R)$$Mg \le \mu (M\omega^2 R)$$ g le mu omega^2 R$$g \le \mu \omega^2 R$$ omega^2 ge fracgmu R$$\omega^2 \ge \frac{g}{\mu R}$$ For minimum angular velocity: omega_textmin = sqrtfracgmu R$$\omega_{\text{min}} = \sqrt{\frac{g}{\mu R}}$$ ### Pattern Recognition In a "rotor" ride setup, gravity is opposed by friction, and the normal force is solely centripetal. Equating mu F_c$\mu F_c$ to mg$mg$ always yields the minimum spin limit. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula v^2 = u^2 + 2as$v^2 = u^2 + 2as$ ### Core Logic

Since the surface is frictionless (implied as no friction coefficient is given), the only force along the incline opposing the upward motion is the component of gravity, mg sin theta$mg \sin \theta$. Acceleration a = -g sin theta$a = -g \sin \theta$. ### Step 1: Apply Kinematic Equation Using the third equation of motion: V^2 = U^2 + 2as$V^2 = U^2 + 2as$ Set final velocity V = 0$V = 0$: 0 = u^2 - 2(g sin theta) S$$0 = u^2 - 2(g \sin \theta) S$$ S = fracu^22g sin theta$$S = \frac{u^2}{2g \sin \theta}$$ ### Pattern Recognition For upward motion on a smooth incline, deceleration is purely g sin theta$g \sin \theta$. The stopping distance is always u^2 / 2a$u^2 / 2a$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

### Related Formula K ell = textconstant$$K \ell = \text{constant}$$ K propto frac1ell$$K \propto \frac{1}{\ell}$$ ### Core Logic

The spring of total length ell$\ell$ is cut into two parts with a length ratio of 1:3. The lengths of the two pieces are: ell_1 = frac14 ell$\ell_1 = \frac{1}{4} \ell$ (smaller piece) ell_2 = frac34 ell$\ell_2 = \frac{3}{4} \ell$ (larger piece) ### Step 1: Calculate Spring Constant of Smaller Piece Since K ell = K' ell'$K \ell = K' \ell'$, for the smaller piece: K ell = K' left(fracell4right)$$K \ell = K' \left(\frac{\ell}{4}\right)$$ K' = 4K$K' = 4K$ Given K = 15 text N/m$K = 15 \text{ N/m}$: K' = 4 times 15 = 60 text N/m$$K' = 4 \times 15 = 60 \text{ N/m}$$ ### Pattern Recognition Spring constant is inversely proportional to spring length. Cutting a spring to 1/n$1/n$ of its length increases its stiffness by a factor of n$n$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula f_k = mu_k N$f_k = \mu_k N$ F_textnet = 0 quad (textfor constant speed)$$F_{\text{net}} = 0 \quad (\text{for constant speed})$$ ### Core Logic

Since C moves with constant speed, a=0$a=0$, meaning net force is zero. For block A (4text kg$4\text{ kg}$) over B, sliding friction f_A = mu (m_A)g = 0.5 times 4 times 10 = 20text N$f_A = \mu (m_A)g = 0.5 \times 4 \times 10 = 20\text{ N}$. For block B (6text kg$6\text{ kg}$) over C, the normal force includes block A. N_B = (m_A + m_B)g = 100text N$N_B = (m_A + m_B)g = 100\text{ N}$. Friction on B from C is f_B = 0.5 times 100 = 50text N$f_B = 0.5 \times 100 = 50\text{ N}$. For block C (8text kg$8\text{ kg}$) over ground, normal force is total weight N_C = (m_A + m_B + m_C)g = 180text N$N_C = (m_A + m_B + m_C)g = 180\text{ N}$. Friction on C from ground is f_C = 0.5 times 180 = 90text N$f_C = 0.5 \times 180 = 90\text{ N}$. ### Step 1: Force Balance Equations For block B to move relative to A and C (assuming standard string configuration where pulling C left causes B to be dragged right relative to C via pulley): The tension T$T$ on B balances friction from A and C: T = f_A + f_B = 20 + 50 = 70text N$$T = f_A + f_B = 20 + 50 = 70\text{ N}$$ For block C moving left, the force F$F$ must overcome friction from ground, friction from B, and the tension T$T$ attached to it (depending on pulley setup. If pulley is on wall, string pulls C right): Assuming standard setup where string connects B and C around wall pulley: F = f_C + f_B + T$$F = f_C + f_B + T$$ F = 90 + 50 + 70 = 210text N$$F = 90 + 50 + 70 = 210\text{ N}$$ ### Pattern Recognition In multi-block pulley systems, pulling the bottom block accumulates all nested friction forces. The bottom block fights the ground, the block above it, AND the tension of the string anchoring the block above it. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion