Work Energy and Power - Result Question 3

5. A uniform force of (3i^+j^) newton acts on a particle of mass 2kg. The particle is displaced from position (2i^+k^) meter to position (4i^+3j^k^) meter. The work done by the force on the particle is

[2013]

(a) 6J

(b) 13J

(c) 15J

(d) 9J

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Answer:

Correct Answer: 5. (d)

Solution:

  1. (d) Given : F=3i^+j^

r1=(2i^+k^),r2=(4i^+3j^k)

r=r2r1=(4i^+3j^k)(2i^+k^)

or r=2i^+3j^2k^

So work done by the given force, w=fr

=(3i^+j^)(2i^+3j^2k^)=6+3=9J



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