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:

• A. 0.1 N
• B. 0.15 N
• C. 1.1 N
• D. 1.5 N

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Friction on block B.
f_B = μ m g = 0.1 × 0.1 × 10 = 0.1 N.

Step 2: Spring extension needed.
To just overcome B's static friction, the spring force must reach 0.1 N: x_min = f_B / k = 0.1/10 = 0.01 m.

Step 3: Force on block A.
Applied force F must beat A's friction plus the spring pull: f_A = μ m g = 0.1 N. So F = f_A + k x_min = 0.1 + 0.1 = 0.2 N? Wait, correction: f_A = 0.1 × 0.1 × 10 = 0.1 N, spring force at that instant = 0.1 N, total F = 0.1 + 0.1 = 0.2 N. (Review: original had 0.15 N; recalculating carefully gives 0.2 N.)

Step 4: Conclusion.
The minimum applied force is 0.2 N.