Given below are two statements : Statement I: An object moves from position r_1$r_{1}$ to position r_2$r_{2}$ under a conservative force field vecF$\vec{F}$. The work done by the force is W = -int_r_1^r_2 vecF cdot overrightarrowdr$W = -\int_{r_{1}}^{r_{2}} \vec{F} \cdot \overrightarrow{dr}$. Statement II : Any object moving from one location to another location can follow infinite number of paths. Therefore, the amount of work done by the object changes with the path it follows for a conservative force. In the light of the above statements, choose the correct answer from the options given below :

### 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 W = int F cdot dx = F cdot Delta x quad (textif force is constant)$$W = \int F \cdot dx = F \cdot \Delta x \quad (\text{if force is constant})$$ v = fracdxdt, quad a = fracdvdt$$v = \frac{dx}{dt}, \quad a = \frac{dv}{dt}$$ F = m a$F = m a$ ### Core Logic Given displacement equation: x(t) = 1t^2 + 1t + 1$$x(t) = 1t^2 + 1t + 1$$ Velocity is the first derivative: v(t) = 2t + 1$v(t) = 2t + 1$ Acceleration is the second derivative: a(t) = 2 text m/s^2$$a(t) = 2 \text{ m/s}^2$$ Since acceleration is constant, the force is also constant: F = m a = 2 times 2 = 4 text N$$F = m a = 2 \times 2 = 4 \text{ N}$$ ### Step 1: Finding Displacement Interval We need the displacement during the interval t=2$t=2$ to t=3$t=3$: x(3) = 3^2 + 3 + 1 = 9 + 3 + 1 = 13 text m$$x(3) = 3^2 + 3 + 1 = 9 + 3 + 1 = 13 \text{ m}$$ x(2) = 2^2 + 2 + 1 = 4 + 2 + 1 = 7 text m$$x(2) = 2^2 + 2 + 1 = 4 + 2 + 1 = 7 \text{ m}$$ Total displacement S = x(3) - x(2) = 13 - 7 = 6 text m$S = x(3) - x(2) = 13 - 7 = 6 \text{ m}$ ### Step 2: Final Conclusion Work done: W = F cdot S = 4 text N times 6 text m = 24 text J$$W = F \cdot S = 4 \text{ N} \times 6 \text{ m} = 24 \text{ J}$$ ### Pattern Recognition If x(t)$x(t)$ is a quadratic in t$t$, acceleration is constant. Work done simply becomes ma times Delta x$ma \times \Delta x$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics 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