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Work Energy and Power question

2013 · Q147
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Work Energy and Power question

2013 · Q147

NEETPhysicsWork Energy and PowerMCQ+4 / −1
A uniform force of (3i^+j^)\left( {3\widehat i + \widehat j} \right)(3i+j​) newton acts on a particle of mass 2 kg. Hence the particle is displaced from position (2i^+k^)\left( {2\widehat i + \widehat k} \right)(2i+k) metre to position (4i^+3j^−k^)\left( {4\widehat i + 3\widehat j - \widehat k} \right)(4i+3j​−k) metre. The work done by the force on the particle is
  1. A
    13 J
  2. B
    15 J
  3. C
    9 J
  4. D
    6 J
View written solutionFree

Correct answer: C

Here, F→=(3i^+j^)N\overrightarrow F = \left( {3\widehat i + \widehat j} \right)NF=(3i+j​)N

Initial position, r1→=(2i^+k^)m\overrightarrow {{r_1}} = \left( {2\widehat i + \widehat k} \right)mr1​​=(2i+k)m

Final position, r2→=(4i^+3j^−k^)m\overrightarrow {{r_2}} = \left( {4\widehat i + 3\widehat j - \widehat k} \right)mr2​​=(4i+3j​−k)m

Displacement, r→=r2→−r1→\overrightarrow r = \overrightarrow {{r_2}} - \overrightarrow {{r_1}} r=r2​​−r1​​

r→=(4i^+3j^−k^)m−(2i^+k^)m=2i^+3j^−2k^ m\overrightarrow r = \left( {4\widehat i + 3\widehat j - \widehat k} \right)m - \left( {2\widehat i + \widehat k} \right)m = 2\widehat i + 3\widehat j - 2\widehat k\,mr=(4i+3j​−k)m−(2i+k)m=2i+3j​−2km

Work done, W=F→.r→=(3i^+j^).(2i^+3j^−2k^)=6+3=9 JW = \overrightarrow F .\overrightarrow r = \left( {3\widehat i + \widehat j} \right).\left( {2\widehat i + 3\widehat j - 2\widehat k} \right) = 6 + 3 = 9\,JW=F.r=(3i+j​).(2i+3j​−2k)=6+3=9J

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