Exams › JEE Advanced › Physics
A block of mass m1 rests on a horizontal surface with coefficient of friction mu1. It is connected to block m2 (also on the ground with coefficient of friction mu2) by a light spring of spring constant k. A constant horizontal force F is applied to m1 (pushing it toward m2). Find the maximum compression of the spring just before block m2 begins to slide.
- (mu2 * m2 * g) / k
- (mu1 * m1 * g + mu2 * m2 * g) / k
- (F - mu1 * m1 * g) / k
- (2 * F - (mu1 * m1 + mu2 * m2) * g) / k
Correct answer: (mu2 * m2 * g) / k
Solution
The spring starts to push m2 as m1 is pushed by F. Block m2 remains stationary as long as the spring force is less than the maximum static friction on m2, which is mu2 * m2 * g. The maximum compression of the spring just before m2 slides is when spring force = mu2 * m2 * g. Therefore k * x = mu2 * m2 * g, so x_max = mu2 * m2 * g / k. Note: this is the compression at which m2 is about to slide — it is determined solely by m2's friction and is independent of F and mu1 (those determine when the dynamics unfold, not the critical compression value).
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