Question:
Two bars A and B each of mass 100 g connected by a light spring of spring constant 10 N/m rest on a horizontal plane without tension in the string. The coefficient of friction between these bars and the surface is 0.1. The minimum force to be applied in the horizontal direction to bar A in order to shift bar B is:
Two bars A and B each of mass 100 g connected by a light spring of spring constant 10 N/m rest on a horizontal plane without tension in the string. The coefficient of friction between these bars and the surface is 0.1. The minimum force to be applied in the horizontal direction to bar A in order to shift bar B is:
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When two objects are connected by a spring, the minimum applied force must overcome both friction and the spring restoring force to move the second block.
Updated On: Jun 19, 2026
- 0.1 N
- 0.15 N
- 1.1 N
- 1.5 N
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the problem.
We have two blocks connected by a spring with spring constant \(k = 10~\text{N/m}\), each of mass \(m = 0.1~\text{kg}\), resting on a surface with coefficient of friction \(\mu = 0.1\). The minimum force \(F\) applied on bar A should be sufficient to overcome the friction on bar B and stretch the spring slightly.
Step 2: Frictional force on bar B.
\[ f_B = \mu m g = 0.1 \cdot 0.1 \cdot 10 = 0.1~\text{N} \]
Step 3: Spring force requirement.
The spring will stretch just enough to overcome the friction on B. For minimal force, consider maximum static friction just overcome by spring: \[ F_\text{spring} = k x_\text{min} = f_B \] \[ x_\text{min} = \frac{f_B}{k} = \frac{0.1}{10} = 0.01~\text{m} \]
Step 4: Force on A.
The applied force on A must overcome its own friction and the spring reaction: \[ F = f_A + k x_\text{min} = 0.1 \cdot 0.1 \cdot 10 + 0.1 = 0.1 + 0.05 = 0.15~\text{N} \]
Step 5: Conclusion.
Thus, the minimum force to shift bar B is 0.15 N.
We have two blocks connected by a spring with spring constant \(k = 10~\text{N/m}\), each of mass \(m = 0.1~\text{kg}\), resting on a surface with coefficient of friction \(\mu = 0.1\). The minimum force \(F\) applied on bar A should be sufficient to overcome the friction on bar B and stretch the spring slightly.
Step 2: Frictional force on bar B.
\[ f_B = \mu m g = 0.1 \cdot 0.1 \cdot 10 = 0.1~\text{N} \]
Step 3: Spring force requirement.
The spring will stretch just enough to overcome the friction on B. For minimal force, consider maximum static friction just overcome by spring: \[ F_\text{spring} = k x_\text{min} = f_B \] \[ x_\text{min} = \frac{f_B}{k} = \frac{0.1}{10} = 0.01~\text{m} \]
Step 4: Force on A.
The applied force on A must overcome its own friction and the spring reaction: \[ F = f_A + k x_\text{min} = 0.1 \cdot 0.1 \cdot 10 + 0.1 = 0.1 + 0.05 = 0.15~\text{N} \]
Step 5: Conclusion.
Thus, the minimum force to shift bar B is 0.15 N.
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