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Work Power and Energy question

2021 · 22 Jul · Shift 2 · Q50
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  5. /2021 · 22 Jul · Shift 2 · Q50

Work Power and Energy question

2021 · 22 Jul · Shift 2 · Q50

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A porter lifts a heavy suitcase of mass 80 kg and at the destination lowers it down by a distance of 80 cm with a constant velocity. Calculate the work done by the porter in lowering the suitcase. (take g = 9.8 ms −-− 2)
  1. A
    +627.2 J
  2. B
    −-− 62720.0 J
  3. C
    −-− 627.2 J
  4. D
    784.0 J
View written solutionFree

Correct answer: C

  1. Given data

    • Mass of suitcase: m=80 kgm = 80\,\text{kg}m=80kg
    • Distance lowered: s=80 cm=0.8 ms = 80\,\text{cm} = 0.8\,\text{m}s=80cm=0.8m
    • Acceleration due to gravity: g=9.8 m s−2g = 9.8\,\text{m s}^{-2}g=9.8m s−2

  2. Find the forces involved

    • Weight of suitcase acts downward: mg=80×9.8=784 Nmg = 80 \times 9.8 = 784\,\text{N}mg=80×9.8=784N
    • Since the suitcase is lowered with constant velocity, acceleration is zero.
    • Therefore, net force is zero, so the porter must apply an upward force equal to 784 N784\,\text{N}784N.

  3. Direction of force and displacement

    • Force applied by porter is upward.
    • Displacement of suitcase is downward.
    • Hence, angle between force and displacement is 180∘180^\circ180∘.

  4. Work done by the porter W=Fscos⁡θW = Fs\cos\thetaW=Fscosθ W=784×0.8×cos⁡180∘W = 784 \times 0.8 \times \cos 180^\circW=784×0.8×cos180∘ W=784×0.8×(−1)W = 784 \times 0.8 \times (-1)W=784×0.8×(−1) W=−627.2 JW = -627.2\,\text{J}W=−627.2J

  5. Match with options The correct option is: C: −627.2 J\boxed{\text{C: } -627.2\,\text{J}}C: −627.2J​

  6. Comparison with stored answer

    • Stored correct answer: C
    • Derived answer: C
    • They match.
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