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JEE Main 2016 (10 Apr, Shift 1) · Work-Energy Theorem

Work-Energy Theorem question from JEE Main 2016 (10 Apr, Shift 1)

Q23JEE Main 2016 · 10 APR · SHIFT 1Medium

Work-Energy Theorem

Velocity-time graph for a body of mass 10 kg is shown in figure. Work-done on the body in first two seconds of the motion is:

Figure

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Answer: (c)

  1. From the graph: v=50−5tv = 50 - 5t (falls from 5050 to 00 in 10 s10\text{ s}).
  2. v(0)=50 m/sv(0) = 50\text{ m/s}, v(2)=40 m/sv(2) = 40\text{ m/s}
  3. W=ΔK=12(10)(402−502)=5×(−900)=−4500 JW = \Delta K = \frac{1}{2}(10)(40^2 - 50^2) = 5 \times (-900) = -4500\text{ J}
  4. So the answer is (c) −4500 J-4500\text{ J}.
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