### Related Formula a = vfracdvdx$a = v\frac{dv}{dx}$ F = m cdot a$F = m \cdot a$ [cite: 778] ### Core Logic Given velocity as a function of position x$x$: [cite: 164, 779] v = 4sqrtx implies v^2 = 16x$v = 4\sqrt{x} \implies v^2 = 16x$ [cite: 164, 779] Differentiating both sides with respect to position coordinate x$x$: [cite: 780] 2vfracdvdx = 16 implies vfracdvdx = 8$2v\frac{dv}{dx} = 16 \implies v\frac{dv}{dx} = 8$ [cite: 780, 781] Thus, the acceleration of the object is a constant value a = 8\ textm/s^2$a = 8\ \text{m/s}^2$[cite: 781]. Converting mass to kilograms (m = 500\ textg = 0.5\ textkg$m = 500\ \text{g} = 0.5\ \text{kg}$) [cite: 164, 788]: F = 0.5 times 8 = 4\ textN$F = 0.5 \times 8 = 4\ \text{N}$ [cite: 788] ### Pattern Recognition When velocity depends on position coordinate x$x$ like v = ksqrtx$v = k\sqrt{x}$, squaring instantly reveals that acceleration is constant, since v^2 = k^2 x$v^2 = k^2 x$ matches the third kinematic profile v^2 = 2ax$v^2 = 2ax$ directly[cite: 779]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula The work done W$W$ by a variable force component F(x)$F(x)$ over a displacement step is given by: W = int_x_1^x_2 F(x) dx$W = \int_{x_{1}}^{x_{2}} F(x) dx$ ### Core Logic Assuming the object moves from the origin x=0$x=0$ to x=1text m$x=1\text{ m}$ : W = int_0^1 (alpha + beta x^2) dx = 5text J$W = \int_{0}^{1} (\alpha + \beta x^{2}) dx = 5\text{ J}$ ### Step 1: Integration and Variable Isolation Perform the integration step : W = left[ alpha x + fracbeta x^33 ight]_0^1 = alpha + fracbeta3 = 5$W = \left[ \alpha x + \frac{\beta x^{3}}{3} ight]_{0}^{1} = \alpha + \frac{\beta}{3} = 5$ Given alpha = 1text N$\alpha = 1\text{ N}$ , substitute this into the equation to find beta$\beta$ : 1 + fracbeta3 = 5 implies fracbeta3 = 4 implies beta = 12text N/m^2$1 + \frac{\beta}{3} = 5 \implies \frac{\beta}{3} = 4 \implies \beta = 12\text{ N/m}^{2}$ ### Pattern Recognition For polynomial forces, the integration steps always yield fractional coefficients matching their power index (1$1$ for constant, frac13$\frac{1}{3}$ for squared terms). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula * **Average Power** (P_textavg$P_{\text{avg}}$) = fractextTotal Work DonetextTotal Time = fracvecF cdot vecst$\frac{\text{Total Work Done}}{\text{Total Time}} = \frac{\vec{F} \cdot \vec{s}}{t}$ * **Instantaneous Power** (P_textinst$P_{\text{inst}}$) = vecF cdot vecv(t)$\vec{F} \cdot \vec{v}(t)$ ### Core Logic Given parameters: * Force vector: vecF = 2hati + 3hatj$\vec{F} = 2\hat{i} + 3\hat{j}$ * Displacement coordinates: P(3,4) rightarrow Q(6,10) implies vecs = (6-3)hati + (10-4)hatj = 3hati + 6hatj$P(3,4) \rightarrow Q(6,10) \implies \vec{s} = (6-3)\hat{i} + (10-4)\hat{j} = 3\hat{i} + 6\hat{j}$ * Time window, t = 4text s$t = 4\text{ s}$ Calculate Average Power : W = vecF cdot vecs = (2hati + 3hatj) cdot (3hati + 6hatj) = (2 times 3) + (3 times 6) = 6 + 18 = 24 text J$W = \vec{F} \cdot \vec{s} = (2\hat{i} + 3\hat{j}) \cdot (3\hat{i} + 6\hat{j}) = (2 \times 3) + (3 \times 6) = 6 + 18 = 24 \text{ J}$ P_textavg = fracWt = frac244 = 6 text W$P_{\text{avg}} = \frac{W}{t} = \frac{24}{4} = 6 \text{ W}$ Now, analyze the instantaneous dynamics to extract final velocity vecv$\vec{v}$ at t=4text s$t=4\text{ s}$: Acceleration vector : veca = fracvecFm = frac2hati + 3hatj4 = 0.5hati + 0.75hatj$\vec{a} = \frac{\vec{F}}{m} = \frac{2\hat{i} + 3\hat{j}}{4} = 0.5\hat{i} + 0.75\hat{j}$ Assuming the body starts from rest, velocity at t=4text s$t=4\text{ s}$ is : vecv = veca cdot t = (0.5hati + 0.75hatj) times 4 = 2hati + 3hatj$\vec{v} = \vec{a} \cdot t = (0.5\hat{i} + 0.75\hat{j}) \times 4 = 2\hat{i} + 3\hat{j}$ Calculate Instantaneous Power at t=4text s$t=4\text{ s}$ : P_textinst = vecF cdot vecv = (2hati + 3hatj) cdot (2hati + 3hatj) = 2^2 + 3^2 = 4 + 9 = 13 text W$P_{\text{inst}} = \vec{F} \cdot \vec{v} = (2\hat{i} + 3\hat{j}) \cdot (2\hat{i} + 3\hat{j}) = 2^2 + 3^2 = 4 + 9 = 13 \text{ W}$ Taking the final ratio : fracP_textavgP_textinst = frac613$\frac{P_{\text{avg}}}{P_{\text{inst}}} = \frac{6}{13}$ *(Note: There is a minor kinematic inconsistency in the question data layout regarding matching coordinate parameters independently, but the calculations follow the standard intended framework directly).* ### Pattern Recognition For constant force acceleration from rest, average power equals frac12 F a t$\frac{1}{2} F a t$ while instantaneous power scales linearly as F a t$F a t$, meaning the structural ratio simplifies exactly to 1:2$1:2$. The custom displacement vector here alters that baseline baseline ratio as tracked. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula V_textBottom = sqrtV_textTop^2 + 4gR$V_{\text{Bottom}} = \sqrt{V_{\text{Top}}^2 + 4gR}$ ### Core Logic Given velocity at the top position : V_textTop = sqrtn^2 gR$V_{\text{Top}} = \sqrt{n^2 gR}$ By work-energy theorem, mechanical energy conservation between the top and bottom positions gives : V_textBottom = sqrtn^2 gR + 4gR$V_{\text{Bottom}} = \sqrt{n^2 gR + 4gR}$ Since KE = frac12m v^2$KE = \frac{1}{2}m v^2$, the ratio of kinetic energy at the bottom to that at the top is [cite: 2, 112]: textRatio = fracV_textBottom^2V_textTop^2 = fracn^2 gR + 4gRn^2 gR = fracn^2 + 4n^2$\text{Ratio} = \frac{V_{\text{Bottom}}^2}{V_{\text{Top}}^2} = \frac{n^2 gR + 4gR}{n^2 gR} = \frac{n^2 + 4}{n^2}$ ### Pattern Recognition Kinetic energy change in vertical circles always gains a fixed additive value of 2mg(2R) = 4mgR$2mg(2R) = 4mgR$ due to gravity work[cite: 112, 687]. ### Chapter Mix Class 11 Physics: Work, Energy and Power