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

Step 1: Understanding the Concept:
The motion of the fireman is governed by Newton's Second Law: \( F_{net} = ma \).
Two forces are acting: weight downward and friction upward.
: Key Formula or Approach:
\[ mg - f = ma \]
Step 2: Detailed Explanation:
Given data:
Mass \( m = 80 \text{ kg} \).
Resisting force \( f = 720 \text{ N} \).
Acceleration due to gravity \( g = 10 \text{ m/s}^2 \).
Calculate weight \( W = mg = 80 \times 10 = 800 \text{ N} \).
Net downward force \( F_{net} = W - f = 800 - 720 = 80 \text{ N} \).
Using \( F = ma \):
\[ 80 = 80 \times a \]
\[ a = \frac{80}{80} = 1 \text{ m/s}^2 \]
Step 3: Final Answer:
The acceleration is \( 1 \text{ m/s}^2 \).