A particle is released from height S$S$ above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

### Related Formula F = -fracdVdx$$F = -\frac{dV}{dx}$$ ### Core Logic The magnitude of the force is given by the absolute value of the slope of the potential energy versus position curve: |F| = |textslope of V-x graph|$|F| = |\text{slope of V-x graph}|$. Comparing the slopes of different regions from the graph: - Region BC has the steepest slope (largest magnitude). - Region AB has a moderate positive slope. - Region DE has a moderate negative slope, but less steep than AB. - Region CD is perfectly flat (slope = 0). ### Step 1: Ranking Forces Therefore, magnitudes of the force follow the order: F_BC > F_AB > F_DE > F_CD$$F_{BC} > F_{AB} > F_{DE} > F_{CD}$$ ### Pattern Recognition In a V-x$V-x$ graph, steeper slopes always mean a stronger restoring or repulsive force. A flat line means zero force (equilibrium). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula By conservation of linear momentum, the common velocity v_textcm$v_{\text{cm}}$ of the combined mass system is: m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_textcm$$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{\text{cm}}$$ By conservation of energy, the loss in Kinetic Energy during maximum compression converted to the potential energy of the spring: Delta K = frac12 k x^2 Rightarrow frac12 m_1 v^2 - frac12 (m_1 + m_2) v_textcm^2 = frac12 k x^2$$\Delta K = \frac{1}{2} k x^2 \Rightarrow \frac{1}{2} m_1 v^2 - \frac{1}{2} (m_1 + m_2) v_{\text{cm}}^2 = \frac{1}{2} k x^2$$ ### Core Logic Given parameters: - m_1 = 10mathrm~kg$m_1 = 10\mathrm{~kg}$, m_2 = 5mathrm~kg$m_2 = 5\mathrm{~kg}$ - Initial velocity of A: v = 3mathrm~m/s$v = 3\mathrm{~m/s}$ - Initial velocity of B: v_2 = 0$v_2 = 0$ - Spring constant k = 3000mathrm~N/m$k = 3000\mathrm{~N/m}$ ### Step 1: Calculate the common center of mass velocity (v_textcm$v_{\text{cm}}$) v_textcm = frac10 times 3 + 5 times 010 + 5 = frac3015 = 2mathrm~m/s$$v_{\text{cm}} = \frac{10 \times 3 + 5 \times 0}{10 + 5} = \frac{30}{15} = 2\mathrm{~m/s}$$ ### Step 2: Apply Energy Conservation to find spring compression ($ ### Step 2: Apply Energy Conservation to find spring compression ($x) $) $frac12 k x^2 = K_i - K_f$\frac{1}{2} k x^2 = K_i - K_f$ $ $frac12 (3000) x^2 = left[ frac12 (10) (3^2) right] - left[ frac12 (10 + 5) (2^2) right]$\frac{1}{2} (3000) x^2 = \left[ \frac{1}{2} (10) (3^2) \right] - \left[ \frac{1}{2} (10 + 5) (2^2) \right]$ $ $1500 x^2 = frac12(90) - frac12(15)(4)$1500 x^2 = \frac{1}{2}(90) - \frac{1}{2}(15)(4)$ $ $1500 x^2 = 45 - 30 = 15$1500 x^2 = 45 - 30 = 15$ $ $x^2 = frac151500 = frac1100$x^2 = \frac{15}{1500} = \frac{1}{100}$ $ $x = frac110mathrm~m = 0.1mathrm~m$x = \frac{1}{10}\mathrm{~m} = 0.1\mathrm{~m}$ ### Pattern Recognition For maximum compression in block-spring-block collisions, the relative kinetic energy gets fully transformed into spring potential energy. $ ### Pattern Recognition For maximum compression in block-spring-block collisions, the relative kinetic energy gets fully transformed into spring potential energy. $frac12 mu v_textrel^2 = frac12 k x^2$\frac{1}{2} \mu v_{\text{rel}}^2 = \frac{1}{2} k x^2$ where $ where $\mu = \frac{m_1 m_2}{m_1 + m_2} is the reduced mass. Here: $ is the reduced mass. Here: $mu = frac10 times 515 = frac103mathrm~kg$\mu = \frac{10 \times 5}{15} = \frac{10}{3}\mathrm{~kg}$ $ $frac12 left(frac103right) (3)^2 = frac12 (3000) x^2 Rightarrow 15 = 1500 x^2 Rightarrow x = 0.1mathrm~m$\frac{1}{2} \left(\frac{10}{3}\right) (3)^2 = \frac{1}{2} (3000) x^2 \Rightarrow 15 = 1500 x^2 \Rightarrow x = 0.1\mathrm{~m}$$ Using reduced mass simplifies center-of-mass collision problems instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula The electrical input power P_textin$P_{\text{in}}$ delivered to the motor is: P_textin = V cdot I$$P_{\text{in}} = V \cdot I$$ The fraction of power lost is determined by the efficiency eta$\eta$: P_textlost = P_textin - P_textout = P_textin (1 - eta)$$P_{\text{lost}} = P_{\text{in}} - P_{\text{out}} = P_{\text{in}} (1 - \eta)$$ To convert watts (mathrmJ/s$\mathrm{J/s}$) to calories per second: P_textlost (mathrmcal/s) = fracP_textlost (mathrmW)4.2$$P_{\text{lost}} (\mathrm{cal/s}) = \frac{P_{\text{lost}} (\mathrm{W})}{4.2}$$ ### Core Logic Given parameters: - Potential $ ### Core Logic Given parameters: - Potential $V = 100\mathrm{~V} - Current $ - Current $I = 1\mathrm{~A} - Efficiency $ - Efficiency $\eta = 91.6\% = 0.916 ### Step 1: Calculate Input Power ($ ### Step 1: Calculate Input Power ($P_{\text{in}}) $) $P_textin = 100 times 1 = 100mathrm~W$P_{\text{in}} = 100 \times 1 = 100\mathrm{~W}$ ### Step 2: Calculate Power Loss in Watts $ ### Step 2: Calculate Power Loss in Watts $P_textlost = 100 times (1 - 0.916) = 100 times 0.084 = 8.4mathrm~W$P_{\text{lost}} = 100 \times (1 - 0.916) = 100 \times 0.084 = 8.4\mathrm{~W}$ ### Step 3: Convert Power Loss to cal/s $ ### Step 3: Convert Power Loss to cal/s $P_textlost = frac8.44.2 = 2mathrm~cal/s$P_{\text{lost}} = \frac{8.4}{4.2} = 2\mathrm{~cal/s}$ ### Pattern Recognition Efficiency calculations often feature clean conversion rates. Recognizing that $ ### Pattern Recognition Efficiency calculations often feature clean conversion rates. Recognizing that $8.4 is exactly twice $ is exactly twice $4.2$ (the conversion factor for Joules to calories) simplifies the final step. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula By the Work-Energy Theorem, the work done by the retarding force equals the change in kinetic energy: W = Delta K = K_f - K_i$$W = \Delta K = K_f - K_i$$ W = int_x_i^x_f F_r(x) dx = frac12 m v_f^2 - frac12 m v_i^2$$W = \int_{x_i}^{x_f} F_r(x) dx = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$$ ### Core Logic Given parameters: - Mass m = 1mathrm~kg$m = 1\mathrm{~kg}$ - Initial velocity v_i = 10mathrm~m/s$v_i = 10\mathrm{~m/s}$ - Region bounds: x_i = 0.1mathrm~m$x_i = 0.1\mathrm{~m}$, x_f = 1.9mathrm~m$x_f = 1.9\mathrm{~m}$ - Retarding force F_r = -kx = -10xmathrm~N$F_r = -kx = -10x\mathrm{~N}$ ### Step 1: Calculate Work Done by the Retarding Force (W$W$) W = int_0.1^1.9 (-10x) dx = -10 left[ fracx^22 right]_0.1^1.9 = -5 left[ (1.9)^2 - (0.1)^2 right]$$W = \int_{0.1}^{1.9} (-10x) dx = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9} = -5 \left[ (1.9)^2 - (0.1)^2 \right]$$ Using the algebraic identity a^2 - b^2 = (a-b)(a+b)$a^2 - b^2 = (a-b)(a+b)$: (1.9)^2 - (0.1)^2 = (1.9 - 0.1)(1.9 + 0.1) = (1.8)(2.0) = 3.6$$(1.9)^2 - (0.1)^2 = (1.9 - 0.1)(1.9 + 0.1) = (1.8)(2.0) = 3.6$$ W = -5 times 3.6 = -18mathrm~J$$W = -5 \times 3.6 = -18\mathrm{~J}$$ ### Step 2: Solve for final velocity (v_f$v_f$) Apply the Work-Energy Theorem: -18 = frac12 (1) v_f^2 - frac12 (1) (10^2)$$-18 = \frac{1}{2} (1) v_f^2 - \frac{1}{2} (1) (10^2)$$ -18 = 0.5 v_f^2 - 50$$-18 = 0.5 v_f^2 - 50$$ 0.5 v_f^2 = 50 - 18 = 32$$0.5 v_f^2 = 50 - 18 = 32$$ v_f^2 = 64 Rightarrow v_f = 8mathrm~m/s$$v_f^2 = 64 \Rightarrow v_f = 8\mathrm{~m/s}$$ ### Pattern Recognition Notice that integrating a linear force $ ### Pattern Recognition Notice that integrating a linear force $F = -kx yields a potential-energy-like term $ yields a potential-energy-like term $\frac{1}{2}k(x_f^2 - x_i^2). Combining this with the Work-Energy theorem gives $. Combining this with the Work-Energy theorem gives $\frac{1}{2} m v_f^2 + \frac{1}{2} k x_f^2 = \frac{1}{2} m v_i^2 + \frac{1}{2} k x_i^2$, which is identical to conservation of mechanical energy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula Instantaneous power P$P$ generated by a force is given by: P = vecF cdot vecv$$P = \vec{F} \cdot \vec{v}$$ Newton's second law: veca = fracvecFm = fracmathrmdvecvmathrmdt implies vecv = int veca \,mathrmdt$$\vec{a} = \frac{\vec{F}}{m} = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t} \implies \vec{v} = \int \vec{a} \,\mathrm{d}t$$ ### Core Logic Convert mass to SI units: m = 1000 mathrm~g = 1 mathrm~kg$$m = 1000 \mathrm{~g} = 1 \mathrm{~kg}$$ Calculate acceleration: veca = frac2thatmathbfi + 3t^2hatmathbfj1 = 2thatmathbfi + 3t^2hatmathbfj$$\vec{a} = \frac{2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}}{1} = 2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}$$ ### Step 1: Determine Velocity Vector Assuming the object starts from rest at t = 0$t = 0$: vecv = int_0^t (2thatmathbfi + 3t^2hatmathbfj) \,mathrmdt = t^2hatmathbfi + t^3hatmathbfj$$\vec{v} = \int_{0}^{t} (2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}) \,\mathrm{d}t = t^2\hat{\mathbf{i}} + t^3\hat{\mathbf{j}}$$ ### Step 2: Calculate Power Compute the dot product of force and velocity: P = vecF cdot vecv = (2thatmathbfi + 3t^2hatmathbfj) cdot (t^2hatmathbfi + t^3hatmathbfj)$$P = \vec{F} \cdot \vec{v} = (2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}) \cdot (t^2\hat{\mathbf{i}} + t^3\hat{\mathbf{j}})$$ P = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5 mathrm~W$$P = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5 \mathrm{~W}$$ ### Pattern Recognition Sees: Time-dependent force $ ### Pattern Recognition Sees: Time-dependent force $\vec{F} \propto t^n on a 1 kg mass. Shortcut: For $ on a 1 kg mass. Shortcut: For $m=1 kg, velocity is the integral of the force components. Power is the dot product of the force vector and its integral. Since $ kg, velocity is the integral of the force components. Power is the dot product of the force vector and its integral. Since $\int at^n \mathrm{d}t = \frac{a}{n+1} t^{n+1}, power component becomes $, power component becomes $\frac{a^2}{n+1} t^{2n+1}. Here, $. Here, $2^2/2 t^3 + 3^2/3 t^5 = 2t^3 + 3t^5 \mathrm{~W}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power