Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 $20\, m$ isPhysics Question Image

KEAMKEAM 2015Work, Energy and Power

Solution:

As, we know, work done by a variable force, $W_{n_{i} \rightarrow n_{f}}$
$=$ area under the force $-$ displacement curve.
Work $=\Lambda$ rea of $\Delta A B F+\Lambda$ rea of $\Delta B C E F$
+ Area of $\Delta E C D$
image
Work done by the particle as it moves from $n=0$ to $20\, m$,
$W=\frac{1}{2} \times 3 \times 5+10 \times 3+\frac{1}{2} \times 5 \times 3$
$\Rightarrow W=\frac{15}{2}+30+\frac{15}{2}$
$\Rightarrow W=15+30=45\, J$