An object of mass $m = 1 k g$ is attached to a platform of mass $M = 4 k g$ with a spring of spring constant $k = 400 N / m$ . There is no friction between the object and the platform, and the coefficient of static friction between the platform and the ground is $μ = 0.1$ . The object is placed at its equilibrium position, and then given a horizontal velocity $υ$
For what $υ$ will the platform never slip on the ground?

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$υ \leq 0.1 m / s$

$υ \leq 0.2 m / s$

$υ \leq 0.25 m / s$

$υ \leq 0.4 m / s$

answer is C.

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Detailed Solution

The force of the spring on the platform is $F_{s} = k x$ , where $x$ is the displacement of the object from equilibrium. The maximum value of $x$ can be found by conservation of energy,
$\frac{1}{2} m υ^{2} = \frac{1}{2} k x^{2}$
Which gives a maximum spring force $F_{s} = υ \sqrt{k m}$ acting on the platform. The maximum possible friction force on the platform is $(M + m) μ g$ . Setting these two equal, the maximum velocity is
$υ = \frac{(M + m) μ g}{\sqrt{k m}} = 0.25 m / s$