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First of all, I am posting on PF after a LONG time so all these changes are new to me. As a result, I have, without reading prior announcements, posted this on the Advanced Physics forum instead of the other one. (To the moderator: Please shift it to the appropriate forum if you think its in the wrong place. Sorry for the inconvenience).

Okay..the problem:

Two bars of masses [itex]m_{1}[/itex] and [itex]m_{2}[/itex] connected by a light undeformed horizontal spring are lying on a uniformly rough horizontal tabletop, having a coefficient of friction [itex]\mu[/itex]. The minimum force that has to be applied horizontally to the bar of mass [itex]m_{1}[/itex] along the length of the spring in order just to shift the other bar of mass [itex]m_{2}[/itex] is

(A) [itex] F = \mu g ( m_{1} + \frac{1}{2}m_{2})[/itex]

(B) [itex] F = \mu g (m_{1} + m_{2})g[/itex]

(C) [itex] F = \mu g (\frac{m_{1} + m_{2}}{2}) [/itex]

(D) [itex] F = \mu g (\frac{1}{2}m_{1} + m_{2})[/itex]

Force Analysis

Suppose the spring has spring constant k and the elongation in it just when motion of [itex]m_{2}[/itex] is to begin is x, then

[tex] F = kx + \mu m_{1}g[/tex]

[tex] kx = \mu m_{2}g[/itex]

This gives part (B) as the answer.

Energy Analysis

The net force on the mass [itex]m_{1}[/itex] is [itex]F - \mu m_{1} g[/itex]. The work of this force is [itex](F - \mu m_{1}g)x[/itex]. This work is stored as potential energy in the spring (since motion of [itex]m_{1}[/itex] has just started, its velocity is zero...to begin with). So,

[tex]Fx = \mu m_{1}g x + \frac{1}{2}kx^{2}[/tex]

Also,

[tex]kx = \mu m_{2} g[/tex]

solving these two equations give (D) as the answer.

Which of these two methods is correct? Why is it correct? Which is wrong? Why is it wrong?

I'd be grateful if someone could help me with this...the two different answers are confusing me somewhat.

Thanks and cheers,

Vivek

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# Homework Help: A Spring + Two Masses [Force Analysis and Energy Analysis giving different answers]

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