Question

A bicycle and rider have a combined mass of 100 kg traveling at 12 km/hr on a level road. A brake is applied to the rear wheel which has 800 mm in diameter. The pressure on the brake is 80 N and the coefficient of friction is 0.05. The distance covered by the bicycle before it comes to rest will be approximately equal to

• A. \( 136 \text{ m} \)
• B. \( 125 \text{ m} \)
• C. \( 250 \text{ m} \)
• D. \( 68 \text{ m} \)

Hint:

When solving problems involving braking distance, first ensure all units are consistent. Calculate the braking force using the coefficient of friction and normal force. Then, use Newton's second law to find the deceleration. Finally, apply a suitable kinematic equation (e.g., \( v^2 = u^2 + 2as \)) to determine the stopping distance.

Answer 1

The Correct Option is A

Step 1: Convert given values to consistent units (SI units).
Combined mass \( m = 100 \text{ kg} \).
Initial velocity \( v_0 = 12 \text{ km/hr} \).
Convert \( v_0 \) to m/s:
\[ v_0 = 12 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 12 \times \frac{5}{18} \text{ m/s} = \frac{60}{18} \text{ m/s} = \frac{10}{3} \text{ m/s} \approx 3.333 \text{ m/s} \] Diameter of rear wheel \( D = 800 \text{ mm} = 0.8 \text{ m} \).
Radius of rear wheel \( R = D/2 = 0.4 \text{ m} \). (Note: The wheel diameter is typically used to find the tangential force, but here the pressure on the brake shoe is given directly as a force).
Pressure on the brake (This is actually the normal force applied by the brake shoe) \( N_b = 80 \text{ N} \).
Coefficient of friction \( \mu_k = 0.05 \).
Step 2: Calculate the braking force.
The braking force (friction force) \( F_f \) is generated at the brake shoe. \[ F_f = \mu_k \times N_b \] \[ F_f = 0.05 \times 80 \text{ N} = 4 \text{ N} \] This braking force acts at the circumference of the wheel and creates a braking torque. This torque causes a deceleration of the bicycle. Assuming the braking force is directly translated to a retarding force on the bicycle, we consider this as the net external force causing deceleration.
Step 3: Calculate the deceleration of the bicycle.
Using Newton's second law, \( F = ma \): \[ F_f = m a \] \[ 4 \text{ N} = 100 \text{ kg} \times a \] \[ a = \frac{4}{100} \text{ m/s}^2 = 0.04 \text{ m/s}^2 \] This is the deceleration.
Step 4: Calculate the distance covered using kinematic equations.
The bicycle comes to rest, so the final velocity \( v = 0 \).
We use the kinematic equation: \( v^2 = v_0^2 + 2as \).
Since it's deceleration, \( a \) is negative.
\[ 0^2 = \left(\frac{10}{3}\right)^2 + 2(-0.04)s \] \[ 0 = \frac{100}{9} - 0.08s \] \[ 0.08s = \frac{100}{9} \] \[ s = \frac{100}{9 \times 0.08} = \frac{100}{0.72} \] \[ s \approx 138.89 \text{ m} \] Let's recheck the calculation with the actual speed value:
\( v_0 = 12 \text{ km/h} = 12 \times \frac{1000}{3600} \text{ m/s} = \frac{120}{36} = \frac{10}{3} \text{ m/s} \).
\( v_0^2 = \left(\frac{10}{3}\right)^2 = \frac{100}{9} \).
The deceleration \( a = \frac{F_f}{m} = \frac{4 \text{ N}}{100 \text{ kg}} = 0.04 \text{ m/s}^2 \).
Using \( v^2 = u^2 + 2as \):
\( 0^2 = \left(\frac{10}{3}\right)^2 - 2(0.04)s \) \( 0 = \frac{100}{9} - 0.08s \)
\( 0.08s = \frac{100}{9} \)
\( s = \frac{100}{9 \times 0.08} = \frac{100}{0.72} \)
\( s = 138.88... \text{ m} \)
Looking at the options, 136 m is the closest approximation. The final answer is $\boxed{\text{1}}$.