Question:
A body is travelling with \(10\,\text{m s}^{-1}\) on a rough horizontal surface. Its velocity after \(2\,\text{s}\) is \(4\,\text{m s}^{-1}\). The coefficient of kinetic friction between the block and the plane is (acceleration due to gravity \(=10\,\text{m s}^{-2}\))
A body is travelling with \(10\,\text{m s}^{-1}\) on a rough horizontal surface. Its velocity after \(2\,\text{s}\) is \(4\,\text{m s}^{-1}\). The coefficient of kinetic friction between the block and the plane is (acceleration due to gravity \(=10\,\text{m s}^{-2}\))
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For motion on a horizontal rough surface, friction produces retardation:
\[
a=\mu g
\]
Always first calculate acceleration using equations of motion and then compare with \(\mu g\).
Updated On: Jun 22, 2026
- \(0.4\)
- \(0.3\)
- \(0.5\)
- \(0.2\)
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The Correct Option is B
Solution and Explanation
Step 1: Use the equation of motion.
We use the first equation of motion: \[ v=u+at \] Given, \[ u=10\,\text{m s}^{-1} \] \[ v=4\,\text{m s}^{-1} \] \[ t=2\,\text{s} \] Substituting the values, \[ 4=10+a(2) \] \[ 4=10+2a \] \[ 2a=-6 \] \[ a=-3\,\text{m s}^{-2} \] Thus, the retardation produced due to friction is \[ 3\,\text{m s}^{-2} \]
Step 2: Relate acceleration with friction.
For motion on a rough horizontal surface, \[ a=\mu g \] where \[ \mu=\text{coefficient of kinetic friction} \] Given, \[ g=10\,\text{m s}^{-2} \] Therefore, \[ 3=\mu(10) \] \[ \mu=\frac{3}{10} \] \[ \mu=0.3 \]
Step 3: Final conclusion.
Hence, the coefficient of kinetic friction is \[ \boxed{0.3} \]
We use the first equation of motion: \[ v=u+at \] Given, \[ u=10\,\text{m s}^{-1} \] \[ v=4\,\text{m s}^{-1} \] \[ t=2\,\text{s} \] Substituting the values, \[ 4=10+a(2) \] \[ 4=10+2a \] \[ 2a=-6 \] \[ a=-3\,\text{m s}^{-2} \] Thus, the retardation produced due to friction is \[ 3\,\text{m s}^{-2} \]
Step 2: Relate acceleration with friction.
For motion on a rough horizontal surface, \[ a=\mu g \] where \[ \mu=\text{coefficient of kinetic friction} \] Given, \[ g=10\,\text{m s}^{-2} \] Therefore, \[ 3=\mu(10) \] \[ \mu=\frac{3}{10} \] \[ \mu=0.3 \]
Step 3: Final conclusion.
Hence, the coefficient of kinetic friction is \[ \boxed{0.3} \]
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