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AIIMS 201926 May
Exam
AIIMS
Year
2019
Date
26 May
Subject
Physics
Chapter
Work, Power and Energy

AIIMS 2019 · 26 May · Physics · Work, Power and Energy

A block of mass 2 ~kg F(t) is free to move along the x -axis. It is...

A block of mass 2 kgF(t)↑2 \mathrm{~kg} F(t) \uparrow is free to move along the xx-axis. It is at rest and from t=0t=0 onwards it is subjected to a timedependent force F(t)F(t) in the xx direction. The force F(t)F(t) varies with tt as shown in the figure. The kinetic energy of the block after 4.5 seconds is A block of mass 2 ~kg F(t) is free to move along the x -axis. It is at rest and from t=0 onwards it is subjected to a

Options

  1. A4.50 J4.50 \mathrm{~J}
  2. B7.50 J7.50 \mathrm{~J}
  3. C5.06 J5.06 \mathrm{~J}
  4. D14.06 J14.06 \mathrm{~J}

Correct answer

C. 5.06 J5.06 \mathrm{~J}

Step-by-step solution

Change in momentum = area under FF - tt graph Δp=12×4×3−12×1.5×43×1.5[ Force at 4.5 s=43×1.5 N]=6−1.5=4.5 kg m s\begin{aligned} \Delta p & =\frac{1}{2} \times 4 \times 3-\frac{1}{2} \times 1.5 \times \frac{4}{3} \times 1.5 \\ & {\left[\text{ Force at } 4.5 \mathrm{~s}=\frac{4}{3} \times 1.5 \mathrm{~N}\right] } \\ & =6-1.5=4.5 \mathrm{~kg} \mathrm{~m} \mathrm{~s}\end{aligned} As Δp=m(v−u)=mv(∵u=0)∴v=Δpm=4.5 kg m s−12 kg=4.52 ms−1\Delta p=m(v-u)=m v \quad(\because u=0) \therefore \quad v=\frac{\Delta p}{m}=\frac{4.5 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}}{2 \mathrm{~kg}}=\frac{4.5}{2} \mathrm{~ms}^{-1} Kinetic energy of the block after 4.5 s4.5 \mathrm{~s} is K=12mv2=12×2 kg×(4.52 ms−1)2=5.06 JK=\frac{1}{2} m v^2=\frac{1}{2} \times 2 \mathrm{~kg} \times\left(\frac{4.5}{2} \mathrm{~ms}^{-1}\right)^2=5.06 \mathrm{~J}
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