Q88 / 160 · JIPMER 2004 · Physics
The work done in pulling up a block of wood weighing 2 kN for a length of 10 m on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is
- A9.82 kJ
- B89 kJ
- C4.35 kJ
- Correct answerD5.17 kJ
Solution
Here : Weight of block $w=2 \mathrm{kN}$,
Distance $d=10 \mathrm{~m}$
Angle of inclination on the plane $\alpha=15^{\circ}$
The block will be pulled up on a smooth plane
Hence, force of resistance due to inclination
$\begin{aligned} F & =w \sin \alpha=2 \times 10^3 \sin 15^{\circ} \\ & =2 \times 10^3 \times 0.2588 \\ & =0.5176 \mathrm{kN}\end{aligned}$
Now work done,
$\begin{aligned}W=F d & =0.5176 \times 10^3 \times 10 \\& =5.17 \mathrm{kJ}\end{aligned}$


Hence, the correct answer is option: 'D'.