Question:
A body is coming with a velocity of 72 km/h on a rough horizontal surface of coefficient of friction 0.5. If the acceleration due to gravity is $10~m/s²$, find the minimum distance it can be stopped.
A body is coming with a velocity of 72 km/h on a rough horizontal surface of coefficient of friction 0.5. If the acceleration due to gravity is $10~m/s²$, find the minimum distance it can be stopped.
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A body is coming with a velocity of 72 km/h on a rough horizontal surface of coefficient of friction 0.5. If the acceleration due to gravity is $10~m/s
Updated On: Apr 15, 2026
- 400 m
- 40 m
- 0.40 m
- 4 m
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The Correct Option is B
Solution and Explanation
Step 1: Values
$u = 72 \text{ km/h} = 20 \text{ m/s}$. Retardation $a = \mu g = 0.5 \times 10 = 5 \text{ m/s}^2$.
Step 2: Equation
$v^2 = u^2 - 2as$.
Step 3: Calculation
$0 = (20)^2 - 2 \times 5 \times s$. $10s = 400 \Rightarrow s = 40 \text{ m}$.
Final Answer: (B)
$u = 72 \text{ km/h} = 20 \text{ m/s}$. Retardation $a = \mu g = 0.5 \times 10 = 5 \text{ m/s}^2$.
Step 2: Equation
$v^2 = u^2 - 2as$.
Step 3: Calculation
$0 = (20)^2 - 2 \times 5 \times s$. $10s = 400 \Rightarrow s = 40 \text{ m}$.
Final Answer: (B)
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