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NEET PHYSICSMedium

A body of mass 1 kg1 \text{ kg} begins to move under the action of a time-dependent force F⃗=(2ti^+3t2j^) N\vec{F} = (2t\hat{i} + 3t^2\hat{j}) \text{ N}, where i^\hat{i} and j^\hat{j} are unit vectors along the X and Y axes. What power will be developed by the force at time tt?

A

(2t2+4t4) W(2t^2 + 4t^4) \text{ W}

B

(2t3+3t4) W(2t^3 + 3t^4) \text{ W}

C

(2t3+3t5) W(2t^3 + 3t^5) \text{ W}

D

(2t+3t3) W(2t + 3t^3) \text{ W}

Step-by-Step Solution

  1. Identify Acceleration: From Newton's Second Law, acceleration a⃗=F⃗m\vec{a} = \frac{\vec{F}}{m}. Given m=1 kgm = 1 \text{ kg}: a⃗=2ti^+3t2j^1=2ti^+3t2j^\vec{a} = \frac{2t\hat{i} + 3t^2\hat{j}}{1} = 2t\hat{i} + 3t^2\hat{j}
  2. Determine Velocity: Velocity is the time integral of acceleration (vecv=∫a⃗dt\\vec{v} = \int \vec{a} dt). Since the body 'begins to move', initial velocity is zero. v⃗=∫(2ti^+3t2j^)dt=(2t22)i^+(3t33)j^=t2i^+t3j^\vec{v} = \int (2t\hat{i} + 3t^2\hat{j}) dt = \left(\frac{2t^2}{2}\right)\hat{i} + \left(\frac{3t^3}{3}\right)\hat{j} = t^2\hat{i} + t^3\hat{j} (Refer to NCERT Class 11, Chapter 3 for kinematic integration).
  3. Calculate Power: Instantaneous power PP is the dot product of force and velocity (P=F⃗⋅v⃗P = \vec{F} \cdot \vec{v}). P=(2ti^+3t2j^)⋅(t2i^+t3j^)P = (2t\hat{i} + 3t^2\hat{j}) \cdot (t^2\hat{i} + t^3\hat{j}) P=(2t)(t2)+(3t2)(t3)P = (2t)(t^2) + (3t^2)(t^3) P=2t3+3t5 WP = 2t^3 + 3t^5 \text{ W} (Refer to NCERT Class 11, Chapter 6, Section 6.10 for the definition of Power).
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