Question

A force $F_x$ acts on a particle such that its position $x$ changes as shown in the figure.
The work done by the particle as it moves from $x=0$ to $20m$ is

• A. 37.5 J
• B. 10J
• C. 15 J
• D. 22.5 J
• E. 45 J

Answer 1

Answer: (E)

As, we know, work done by a variable force, $W_{n_i\to n_f}$
$=$ area under the force $-$ displacement curve.
Work $=\Lambda$ rea of $\Delta ABF+\Lambda$ rea of $\Delta BCEF$
+ Area of $\Delta ECD$

Work done by the particle as it moves from $n=0$ to $20m$,
$W=\frac{1}{2}\times 3\times 5+10\times 3+\frac{1}{2}\times 5\times 3$
$\Rightarrow W=\frac{15}{2}+30+\frac{15}{2}$
$\Rightarrow W=15+30=45J$