Question:
A block is lying at rest inside a bus. The maximum acceleration of the bus such that the block remain stationary is \((\text{the static friction coefficient}=0.2,\ \text{acceleration due to gravity}=10\,\text{m s}^{-2})\)
A block is lying at rest inside a bus. The maximum acceleration of the bus such that the block remain stationary is \((\text{the static friction coefficient}=0.2,\ \text{acceleration due to gravity}=10\,\text{m s}^{-2})\)
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For a block to remain at rest on an accelerating horizontal surface, static friction must satisfy
\[
ma\leq \mu_s mg.
\]
Hence the maximum acceleration is
\[
a_{\max}=\mu_s g.
\]
Updated On: Jun 18, 2026
- \(1\,\text{m s}^{-2}\)
- \(0.5\,\text{m s}^{-2}\)
- \(2\,\text{cm s}^{-2}\)
- \(2\,\text{m s}^{-2}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understand the physical situation.
A block is at rest inside an accelerating bus. For the block to remain stationary relative to the bus, static friction must provide the required horizontal force.
If the bus accelerates with acceleration \(a\), the block must also accelerate with the same acceleration \(a\).
Step 2: Write the required frictional force.
The horizontal force needed to accelerate the block is \[ F=ma \] This force is supplied by static friction.
Step 3: Write the maximum static friction.
The maximum static friction is \[ f_{\max}=\mu_s N \] Since the block is on a horizontal surface, \[ N=mg \] Therefore, \[ f_{\max}=\mu_s mg \]
Step 4: Apply the condition for no slipping.
For the block to remain stationary relative to the bus, \[ ma\leq \mu_s mg \] Cancelling \(m\), \[ a\leq \mu_s g \] The maximum acceleration is therefore \[ a_{\max}=\mu_s g \]
Step 5: Substitute the given values.
Given, \[ \mu_s=0.2 \] and \[ g=10\,\text{m s}^{-2} \] So, \[ a_{\max}=0.2\times 10 \] \[ a_{\max}=2\,\text{m s}^{-2} \]
Step 6: Final conclusion.
Therefore, the maximum acceleration of the bus is \[ \boxed{2\,\text{m s}^{-2}} \]
A block is at rest inside an accelerating bus. For the block to remain stationary relative to the bus, static friction must provide the required horizontal force.
If the bus accelerates with acceleration \(a\), the block must also accelerate with the same acceleration \(a\).
Step 2: Write the required frictional force.
The horizontal force needed to accelerate the block is \[ F=ma \] This force is supplied by static friction.
Step 3: Write the maximum static friction.
The maximum static friction is \[ f_{\max}=\mu_s N \] Since the block is on a horizontal surface, \[ N=mg \] Therefore, \[ f_{\max}=\mu_s mg \]
Step 4: Apply the condition for no slipping.
For the block to remain stationary relative to the bus, \[ ma\leq \mu_s mg \] Cancelling \(m\), \[ a\leq \mu_s g \] The maximum acceleration is therefore \[ a_{\max}=\mu_s g \]
Step 5: Substitute the given values.
Given, \[ \mu_s=0.2 \] and \[ g=10\,\text{m s}^{-2} \] So, \[ a_{\max}=0.2\times 10 \] \[ a_{\max}=2\,\text{m s}^{-2} \]
Step 6: Final conclusion.
Therefore, the maximum acceleration of the bus is \[ \boxed{2\,\text{m s}^{-2}} \]
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