Question

A force $\overrightarrow{F}$ = $k[k\hat{i}+y\hat{j}]$ , where k is a positive constant, acts on a particle moving in x-y plane starting from the point (3,5), the particle is taken along a straight line to (5, 7). The work done by the force is :

• A. zero
• B. 35k
• C. 20k
• D. 14k

Answer 1

The correct option is C

20k

$W=\int F_x{d}_x+\int F_y{d}_y=k\left({\int }_3^{5}xdx+{\int }_5^{7}ydy\right)=k(\frac{25-9}{2}+\frac{49-25}{2})$
= (8+12)k

= 20k