A body of mass 2 text kg$2 \text{ kg}$ begins to move under the action of a time dependent force given by vecF = (6thati + 6t^2hatj)textN$\vec{F} = (6t\hat{i} + 6t^2\hat{j})\text{N}$. The power developed by the force at the time t$t$ is given by:
A.(6t^4 + 9t^5)textW$(6t^4 + 9t^5)\text{W}$
B.(3t^3 + 6t^5)textW$(3t^3 + 6t^5)\text{W}$
C.(9t^5 + 6t^3)textW$(9t^5 + 6t^3)\text{W}$
D.(9t^3 + 6t^5)textW$(9t^3 + 6t^5)\text{W}$
Solution & Explanation
### 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
Keywords:#power developed by the force#time dependent force#JEE Main 2024 Evening Q48#Work, Energy and Power JEE Main 2024#Power by a Variable Force JEE Main 2024
More Work, Energy and Power Previous-Year Questions — Page 5
Q37jee_main_2024_30_jan_morningConservation of Mechanical Energy
A particle is placed at the point A of a frictionless track ABC as shown in figure. It is gently pushed toward right. The speed of the particle when it reaches the point B is: (Take g = 10 mathrm~m/s^2$g = 10 \mathrm{~m/s^2}$).
A particle on a frictionless track moving from height 1m at point A to 0.5m at point B.
### 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
Q49jee_main_2024_31_jan_morningConservation Of Momentum
An artillery piece of mass M_1$M_1$ fires a shell of mass M_2$M_2$ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is :
A.M_1 / (M_1 + M_2)$M_1 / (M_1 + M_2)$
B.fracM_2M_1$\frac{M_2}{M_1}$
C.M_2 / (M_1 + M_2)$M_2 / (M_1 + M_2)$
D.fracM_1M_2$\frac{M_1}{M_2}$
Solution
### 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
More Work, Energy and Power Questions — jee_main_2024_31_jan_evening
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