Question:
A block of mass \(1\) kg is kept on ice surface. When velocity of \(4\) m/s is given to it, it stops by friction in \(5\) second. The coefficient of friction is (acceleration due to gravity, \(g=10\) m/s\({}^{2}\))
A block of mass \(1\) kg is kept on ice surface. When velocity of \(4\) m/s is given to it, it stops by friction in \(5\) second. The coefficient of friction is (acceleration due to gravity, \(g=10\) m/s\({}^{2}\))
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Find the retardation from the stopping time using the first equation of motion, then equate it to the friction force divided by mass.
Updated On: Nov 12, 2026
- \(0.02\)
- \(0.04\)
- \(0.08\)
- \(0.09\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Friction is the only horizontal force, so it produces a retardation \(a=\mu g\).
Step 2: Find the retardation
Using \(v=u-at\) with \(v=0\), \(u=4\) m/s and \(t=5\) s:
\(a=\frac{4}{5}=0.8\) m/s\(^{2}\).
Step 3: Calculation
\(\mu=\frac{a}{g}=\frac{0.8}{10}=0.08\)
Step 4: Compare the options
The coefficient of friction \(0.08\) matches option (C).
Final Answer:
The coefficient of friction is 0.08. \[ \boxed{\text{(C) }0.08} \]
Friction is the only horizontal force, so it produces a retardation \(a=\mu g\).
Step 2: Find the retardation
Using \(v=u-at\) with \(v=0\), \(u=4\) m/s and \(t=5\) s:
\(a=\frac{4}{5}=0.8\) m/s\(^{2}\).
Step 3: Calculation
\(\mu=\frac{a}{g}=\frac{0.8}{10}=0.08\)
Step 4: Compare the options
The coefficient of friction \(0.08\) matches option (C).
Final Answer:
The coefficient of friction is 0.08. \[ \boxed{\text{(C) }0.08} \]
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