An object of mass 1000 \, textg$1000 \, \text{g}$ experiences a time dependent force vecmathrmF = (2thatmathbfi + 3t^2hatmathbfj)mathrmN$\vec{\mathrm{F}} = (2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}})\mathrm{N}$ . The power generated by the force at time t$t$ is:

### Related Formula The force vector vecF$\vec{F}$ is related to the potential energy U$U$ by the negative gradient of potential energy: vecF = -vecnabla U = -left( fracpartial Upartial xhati + fracpartial Upartial yhatj + fracpartial Upartial zhatk right)$$\vec{F} = -\vec{\nabla} U = -\left( \frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k} \right)$$ Hence, the x-component of force is: F_x = -fracpartial Upartial x$$F_x = -\frac{\partial U}{\partial x}$$ ### Core Logic Given the potential energy function: U = 2x^2 + 3y^3 + 2z$$U = 2x^2 + 3y^3 + 2z$$ Taking the partial derivative with respect to x$x$ (treating y$y$ and z$z$ as constants): fracpartial Upartial x = fracpartialpartial x(2x^2) = 4x$$\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2) = 4x$$ ### Step 1: Substitute Coordinates The x-component of the force is: F_x = -4x$F_x = -4x$ At point P(1, 2, 3) mathrm~m$P(1, 2, 3) \mathrm{~m}$, we substitute x = 1$x = 1$: F_x = -4(1) = -4 mathrm~N$$F_x = -4(1) = -4 \mathrm{~N}$$ Magnitude of the x-component of force is: |F_x| = 4 mathrm~N$$|F_x| = 4 \mathrm{~N}$$ ### Pattern Recognition When asked for a specific component (like x-component), only differentiate partially with respect to that specific variable. The remaining coordinates (y, z$y, z$) act purely as constants and vanish. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula K_i + U_i = K_f + U_f$$K_i + U_i = K_f + U_f$$ frac12 m u^2 + mgh_i = frac12 m v^2 + mgh_f$$\frac{1}{2} m u^2 + mgh_i = \frac{1}{2} m v^2 + mgh_f$$ ### Core Logic Since the track is frictionless, mechanical energy is conserved. We can apply the Principle of Conservation of Mechanical Energy (COME) between point A and point B. ### Step 1: Apply Conservation of Energy At point A (initially pushed gently, u approx 0$u \approx 0$): textKE_A + U_A = textKE_B + U_B$$\text{KE}_A + U_A = \text{KE}_B + U_B$$ 0 + mg(h_A) = frac12 mv^2 + mg(h_B)$$0 + mg(h_A) = \frac{1}{2} mv^2 + mg(h_B)$$ Substitute the given values (h_A = 1 mathrm~m$h_A = 1 \mathrm{~m}$, h_B = 0.5 mathrm~m$h_B = 0.5 \mathrm{~m}$): mg(1) = frac12 mv^2 + mg(0.5)$$mg(1) = \frac{1}{2} mv^2 + mg(0.5)$$ mg(0.5) = frac12 mv^2$$mg(0.5) = \frac{1}{2} mv^2$$ v^2 = 2g(0.5) = g$$v^2 = 2g(0.5) = g$$ ### Step 2: Calculate Velocity Given g = 10 mathrm~m/s^2$g = 10 \mathrm{~m/s^2}$: v = sqrtg = sqrt10 mathrm~m/s$$v = \sqrt{g} = \sqrt{10} \mathrm{~m/s}$$ ### Pattern Recognition For a mass sliding down a frictionless slope, its speed relies only on the vertical height dropped: v = sqrt2gDelta h$v = \sqrt{2g\Delta h}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula vecF = mveca$$\vec{F} = m\vec{a}$$ veca = fracdvecvdt implies vecv = int veca dt$$\vec{a} = \frac{d\vec{v}}{dt} \implies \vec{v} = \int \vec{a} dt$$ P = vecF cdot vecv$$P = \vec{F} \cdot \vec{v}$$ ### Core Logic First find acceleration from force and mass. Then integrate acceleration to find the velocity vector at time t$t$ (starting from rest). Finally, compute the dot product of Force and Velocity to get instantaneous power. ### Step 1: Calculate Acceleration Given vecF = (6thati + 6t^2hatj) text N$\vec{F} = (6t\hat{i} + 6t^2\hat{j}) \text{ N}$ and m = 2 text kg$m = 2 \text{ kg}$. veca = fracvecFm = frac6thati + 6t^2hatj2$$\vec{a} = \frac{\vec{F}}{m} = \frac{6t\hat{i} + 6t^2\hat{j}}{2}$$ veca = (3thati + 3t^2hatj) text m/s^2$$\vec{a} = (3t\hat{i} + 3t^2\hat{j}) \text{ m/s}^2$$ ### Step 2: Calculate Velocity Assuming the body begins to move from rest (at t=0, v=0$t=0, v=0$): vecv = int_0^t veca \, dt = int_0^t (3thati + 3t^2hatj) \, dt$$\vec{v} = \int_0^t \vec{a} \, dt = \int_0^t (3t\hat{i} + 3t^2\hat{j}) \, dt$$ vecv = left( frac3t^22 right)hati + left( frac3t^33 right)hatj$$\vec{v} = \left( \frac{3t^2}{2} \right)\hat{i} + \left( \frac{3t^3}{3} \right)\hat{j}$$ vecv = left( frac3t^22 right)hati + t^3hatj$$\vec{v} = \left( \frac{3t^2}{2} \right)\hat{i} + t^3\hat{j}$$ ### Step 3: Calculate Power P = vecF cdot vecv$$P = \vec{F} \cdot \vec{v}$$ P = (6thati + 6t^2hatj) cdot left(frac3t^22hati + t^3hatjright)$$P = (6t\hat{i} + 6t^2\hat{j}) \cdot \left(\frac{3t^2}{2}\hat{i} + t^3\hat{j}\right)$$ P = (6t) times left(frac3t^22right) + (6t^2) times (t^3)$$P = (6t) \times \left(\frac{3t^2}{2}\right) + (6t^2) \times (t^3)$$ P = 9t^3 + 6t^5 text W$$P = 9t^3 + 6t^5 \text{ W}$$ ### Pattern Recognition When force varies as a polynomial in time t^n$t^n$, acceleration does too. Velocity jumps to t^n+1$t^{n+1}$. Power (F cdot v$F \cdot v$) will result in terms behaving as t^2n+1$t^{2n+1}$. Here t to t^3$t \to t^3$ and t^2 to t^5$t^2 \to t^5$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power Class 11 Physics: Motion in a Plane

### Related Formula textKE = fracp^22m$$\text{KE} = \frac{p^2}{2m}$$ ### Core Logic By conservation of linear momentum (since no external horizontal force acts on the system): 0 = M_1 v_1 + M_2 v_2$$0 = M_1 v_1 + M_2 v_2$$ |vecp_1| = |vecp_2| = p$$|\vec{p_1}| = |\vec{p_2}| = p$$ Both the artillery and the shell acquire the exact same magnitude of momentum during firing. ### Step 2: Kinetic Energy Ratio The kinetic energy is related to momentum by textKE = fracp^22m$\text{KE} = \frac{p^2}{2m}$. Since p$p$ is identical for both bodies: textKE propto frac1m$$\text{KE} \propto \frac{1}{m}$$ Therefore, the ratio of kinetic energy of the artillery (M_1$M_1$) to the shell (M_2$M_2$) is: fractextKE_1textKE_2 = fracfracp^22M_1fracp^22M_2 = fracM_2M_1$$\frac{\text{KE}_1}{\text{KE}_2} = \frac{\frac{p^2}{2M_1}}{\frac{p^2}{2M_2}} = \frac{M_2}{M_1}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy And Power