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Work Power and Energy question

2019 · 10 Jan · Shift 2 · Q61
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Work Power and Energy question

2019 · 10 Jan · Shift 2 · Q61

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A particle which is experiencing a force, given by F→=3i^−12j^,\overrightarrow F = 3\widehat i - 12\widehat j,F=3i−12j​, undergoes a displacement of d→=4i→\overrightarrow d = 4\overrightarrow id=4i particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?
  1. A
    9 J
  2. B
    10 J
  3. C
    12 J
  4. D
    15 J
View written solutionFree

Correct answer: D

  1. Use the work-energy theorem

    The change in kinetic energy is equal to the work done by the force:

    Delta K = W = \vec F \cdot \vec d

  2. Write the given vectors

    F⃗=3i^−12j^\vec F = 3\hat i - 12\hat jF=3i^−12j^​ d⃗=4i^\vec d = 4\hat id=4i^

  3. Find the work done using dot product

    W=(3i^−12j^)⋅(4i^)W = (3\hat i - 12\hat j) \cdot (4\hat i)W=(3i^−12j^​)⋅(4i^)

    Since j^⋅i^=0\hat j \cdot \hat i = 0j^​⋅i^=0 and i^⋅i^=1\hat i \cdot \hat i = 1i^⋅i^=1,

    W=3⋅4+(−12)⋅0=12 JW = 3 \cdot 4 + (-12)\cdot 0 = 12\text{ J}W=3⋅4+(−12)⋅0=12 J

  4. Apply work-energy theorem

    Initial kinetic energy:

    Ki=3 JK_i = 3\text{ J}Ki​=3 J

    Therefore final kinetic energy is

    Kf=Ki+W=3+12=15 JK_f = K_i + W = 3 + 12 = 15\text{ J}Kf​=Ki​+W=3+12=15 J

  5. Match with options

    Kf=15 JK_f = 15\text{ J}Kf​=15 J

    So the correct option is D.

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