What Minimum Force Moves the Bar?

[ambuj123]

Here is this problem which is bugging me:
Q. Two bars of mass m1 and m2 are connected by a non deformed light spring rest on a horizontol plane . coefficient of friction between bars and plane is k . what minimum constant horizontol force has to be applied to bar of mass m1 inorder to move other bar.?

A this is what is was trying to do
at the time bar m2 moves maximum static friction acts and is equal to LX where X is extension of spring and L Hookes constant of spring
for m1 similarly we write force equation in horizontol direction and put LX value from equation for m2.
the we get relation in acceleration of block 2 and f the we apply chain rule and write a as -(dv/dx).(dx/dt) as velocity is decreasing as hookes force increases with extension.integrating both sides we get we get relation for kinetic energy of m1. then taking m1 and m2 + spring as system we by using work energy thworem write change in total mechanical energy bis equal to net work done by external force friction and F.
well by doing this i get answer
F = kg(m1+3/4m2) while correct answer is F= kg(m1+m2/2)
i don't know ehere i am doing wrong
please help?

 

[maverick280857]

Hello Ambuj

Let's assume that m2 is the mass on the left and m1 is the mass to the right of the spring. Hence the spring is sandwiched between the masses. You need to compute the minimum force on m1 in order to cause motion of m2. Hence, you need to find that critical value of F which causes m2 to start moving from rest (this is what I figured from the wording of your problem..please clarify).

Let the spring constant be $$\mu$$

Suppose I apply F on m1, its force equation is

$$F - \mu x - km_{1}g = m_{1}\frac{dv_{1}}{dt}$$

and for m2, it is

$$kx - km_{2}g = m_{2}\frac{dv_{2}}{dt}$$

Oops...the tex isn't working...I wonder why. SO I'll continue this in a while after you reply back. I am trying to figure it out using force equations first and not work energy. However, if I do figure it out using work energy, my equation could be something like

$$\frac{1}{2}m_{1}v_{1}^2 + \frac{1}{2}m_{2}v_{2}^2 + \frac{1}{2}\mu x^{2} =$$Work done by friction+externally applied force.

For the minimum force, I think we can set v2 = 0 if I understand your reasoning correctly...

 

[M98Ranger]

It seems like you have a good understanding of the problem and the concepts involved. However, it is possible that you made a mistake in your calculations or assumptions. I would recommend double checking your equations and steps to see if there are any errors. Additionally, it might be helpful to try approaching the problem from a different angle or using a different method to see if you get the same answer. If you are still having trouble, you could also consult with a teacher or classmate for clarification or assistance. Don't be discouraged, problem solving can be tricky and it's important to keep trying and learning from any mistakes.