A uniform force of $(3 \mathbf{i}+\mathbf{j}) \mathrm{N}$ acts on a particle of mass $2 \mathrm{~kg}$. Hence…

A uniform force of $(3 \mathbf{i}+\mathbf{j}) \mathrm{N}$ acts on a particle of mass $2 \mathrm{~kg}$. Hence the particle is displaced from position $(2 \mathbf{i}+\mathbf{k}) \mathrm{m}$ to position $(4 \mathbf{i}+3 \mathbf{j}-\mathbf{k}) \mathrm{m}$. The work done by the force on the particle is
  1. $9 \mathrm{~J}$
  2. $6 \mathrm{~J}$
  3. $13 \mathrm{~J}$
  4. $15 \mathrm{~J}$

Solution

Given, force $F=3 i+j$
$\begin{array}{l}
r_1=(2 i+k) m \text { and } r_2 =(4 i+3 j-k) m \\
\therefore s=r_2-r_1=(4 i+3 j-k)-(2 i+k) \\
=(2 i+3 j-2 k) m \\
\begin{array}{l}
\therefore W=F \cdot s & =(3 i+j) \cdot(2 i+3 j-2 k) \\
& =3 \times 2+3+0 \\
& =6+3=9 \mathrm{~J}
\end{array} \\
\end{array}$

Asked in: NEET 2013 (All India)

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