Question

A person is holding a bucket by applying a force of $10N$. He moves a horizontal distance of $5m$ and then climbs up a vertical distance of $10m$. The total work done by him is equal to

• A. $50J$
• B. $150J$
• C. $100J$
• D. $200J$

Answer 1

The correct option is C $100J$

Let $\theta$ be the angle between force and displacement vectors. For horizontal motion,

$F=10N,s=5m,\theta =90°$

Work done, W =
$\overrightarrow{F}.\overrightarrow{s}$
Work done,
$W_1=Fscos\theta =10\times 5\times cos90°=0$

For vertical motion, the angle between force and displacement is 0°.

Here,$F=10N,s=\; 10m,\theta =0°$

Work done,$W_2=10\times 10\times \cos \left(0\right)=100J$

Total work done $=$$W_1+W_2=100J$