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Homework Statement
A light spring having a force constant of 125 N/m is used to pull a 9.50 kg sled on a horizontal ice rink. The coefficient of kinetic friction between the sled and the ice is 0.200. The sled has an acceleration of 2.00 [tex]m/s^2[/tex]. By how much does the spring stretch if it pulls on the sled at 30.0 degrees above the horizontal?2. The attempt at a solution
The following is how I solve the problem by myself.
First, the object is only moving along the horizontal level and there is friction. I assuming the total force which pulls on the sled at 30 degree is F and the friction is f. The force affecting the spring is [tex]F_k = F_x = F\cos(30) - f[/tex].
For friction, it should be
[tex]f = \mu (mg - F\sin(30)) = \mu (mg - F_x\tan(30))[/tex]
and from Newton second law
[tex]F_x - f = ma[/tex]
so
[tex]F_x = ma + f[/tex]
gives
[tex]F_x = \frac{ma + \mu mg}{1+\mu\tan(30)}[/tex]
that is, according to Hook's Law
[tex]\Delta x = F_x/k = \cdots [/tex]
I wonder if my calculation is corrected? I think about it for several times, it seems make sense. But one of my friend come to me few minutes ago. He said "the sled is experiencing a total force (along the horizon), which is [tex]F_x - f = ma[/tex], so the spring will also experience the same (total) force. Hence, x = ma/k instead of [tex]F_x[/tex]/k". This seems also make sense, so which one is correct? WHY?
Here I attach the figure of the problem (given by my TA instead of from the text)
http://img697.imageshack.us/img697/4535/figt.jpg
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