### Neet Solved Paper 2013 Question 6

##### Question: A uniform force of $ (3\mathbf{i}+\mathbf{j}) $ N acts on a particle of mass 2 kg. Hence the particle is displaced from position $ (2\mathbf{i}+\mathbf{k}) $ m to position $ (4\mathbf{i}+3\mathbf{j}-\mathbf{k}) $ m. The work done by the force on the particle is

#### Options:

A) 9 J

B) 6 J

C) 13 J

D) 15 J

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

- Given, force F = 3 i + j $ r _1=(2i+k)m $ and $ r _2=(4i+3J-k)m $

$ \therefore $ $ s=r _2-r _1=(4i+3j-k)-(2i+k) $ $ =(2i+3j-2k)m $

$ \therefore $ $ W=F\cdot s=(3i+j)\cdot (2i+3j-2k) $ $ \begin{aligned} & =3\times 2+3+0 \\ & =6+3=9,J \\ \end{aligned} $