Q126 / 170 · Manipal (MET) 2021 · Physics
A fireman weighing 80 kg slides down a pole. If the resisting force of friction is 720 N , his acceleration would be : (take $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)$
- A$0.11 \mathrm{~m} / \mathrm{s}^2$
- B$0.9 \mathrm{~m} / \mathrm{s}^2$
- Correct answerC$1 \mathrm{~m} / \mathrm{s}^2$
- Dzero
Solution
Force of friction $=720 \mathrm{~N}$
If $a$ is the acceleration of fireman, sliding down the pole, then
$\begin{aligned} m g-F & =m a \\ a & =\frac{m g-F}{m} \\ & =\frac{(80 \times 10-720)}{80} \\ & =1 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
Hence, the correct answer is option: 'C'.