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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 xx and yy-axis. What power will be developed by the force at the time tt?

A

(2t2+4t4) W(2t^2+4t^4)\text{ W}

B

(2t3+3t3) W(2t^3+3t^3)\text{ W}

C

(2t3+3t5) W(2t^3+3t^5)\text{ W}

D

(2t3+3t4) W(2t^3+3t^4)\text{ W}

Step-by-Step Solution

  1. Identify Acceleration: According to Newton's Second Law, F⃗=ma⃗\vec{F} = m\vec{a}. Given m=1 kgm = 1\text{ kg}, the acceleration is a⃗=F⃗/1=2ti^+3t2j^\vec{a} = \vec{F}/1 = 2t\hat i + 3t^2\hat j [Class 11 Physics, Ch 4, Sec 4.5, Eq 4.5].
  2. Calculate Velocity: Velocity is the time integral of acceleration. Assuming the body starts from rest (v=0v=0 at t=0t=0): v⃗=∫a⃗dt=∫(2ti^+3t2j^)dt\vec{v} = \int \vec{a} dt = \int (2t\hat i + 3t^2\hat j) dt v⃗=t2i^+t3j^\vec{v} = t^2\hat i + t^3\hat j
  3. Calculate Power: Instantaneous power (PP) is the dot product of force and velocity. P=F⃗⋅v⃗P = \vec{F} \cdot \vec{v} [Class 11 Physics, Ch 5, Sec 5.10, Eq 5.21] 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)=2t3+3t5 WP = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5\text{ W}
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