Question

A fireman weighing \(80\,\text{kg}\) slides down a pole. If the resisting force of friction is \(720\,\text{N}\), his acceleration would be: (take \( g = 10\,\text{m/s}^2 \))

• A. \(0.11 \, \text{m/s}^2\)
• B. \(0.9 \, \text{m/s}^2\)
• C. \(1 \, \text{m/s}^2\)
• D. zero

Hint:

If friction is close to \(mg\), motion is slow. If equal → zero acceleration.

Answer 1

The Correct Option is C

Concept: Net force acting downward: \[ F_{\text{net}} = mg - f \] \[ a = \frac{F_{\text{net}}}{m} \]

Step 1: Weight of fireman \[ mg = 80 \times 10 = 800 \, \text{N} \]

Step 2: Net force (downward) \[ F_{\text{net}} = 800 - 720 = 80 \, \text{N} \]

Step 3: Acceleration \[ a = \frac{80}{80} = 1 \, \text{m/s}^2 \] Final: \[ {1 \, \text{m/s}^2} \]