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Work Power and Energy question

2004 · Shift 0 · Q159
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Work Power and Energy question

2004 · Shift 0 · Q159

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A force F→=(5i→+3j→+2k→)N\overrightarrow F = \left( {5\overrightarrow i + 3\overrightarrow j + 2\overrightarrow k } \right)NF=(5i+3j​+2k)N is applied over a particle which displaces it from its origin to the point r→=(2i→−j→)m.\overrightarrow r = \left( {2\overrightarrow i - \overrightarrow j } \right)m.r=(2i−j​)m. The work done on the particle in joules is
  1. A
    +10+10+10
  2. B
    +7+7+7
  3. C
    −7-7−7
  4. D
    +13+13+13
View written solutionFree

Correct answer: B

  1. Use the definition of work done by a constant force

For a constant force, work done is the dot product of force and displacement:

W=F⃗⋅r⃗W = \vec F \cdot \vec rW=F⋅r

  1. Write the given vectors

F⃗=(5i^+3j^+2k^) N\vec F = (5\hat i + 3\hat j + 2\hat k)\,\text{N}F=(5i^+3j^​+2k^)N

r⃗=(2i^−j^) m=(2i^−1j^+0k^) m\vec r = (2\hat i - \hat j)\,\text{m} = (2\hat i - 1\hat j + 0\hat k)\,\text{m}r=(2i^−j^​)m=(2i^−1j^​+0k^)m

  1. Compute the dot product

W=(5)(2)+(3)(−1)+(2)(0)W = (5)(2) + (3)(-1) + (2)(0)W=(5)(2)+(3)(−1)+(2)(0)

W=10−3+0=7 JW = 10 - 3 + 0 = 7\,\text{J}W=10−3+0=7J

  1. Match with the options

The work done is:

+7 J\boxed{+7\,\text{J}}+7J​

So, the correct option is B.

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