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Potential energy in spring

1 vote

In the figure shown, pulley and spring are ideal. Find the potential energy stored in the spring (m1 > m2).

I tried as
Let the extension in spring be x
Then m$_1$ moves down by 2x .
By energy conservation m$_1gx=\frac{1}{2}kx^2$
Hence x= (4mg)/(k)
So PE is $\frac{8m^2g^2}{k}$

But the answer given as $\frac{2m^2g^2}{k}$

asked Jan 27, 2017 in Physics Problems by koolman (4,286 points)

1 Answer

1 vote
Best answer

If $m_2$ is attached to the ground then it doesn't really matter what value it has, provided that it is less than $m_1$. The string or wire which attaches $m_2$ to the ground will supply enough force to counteract the weight of $m_1$. (If $m_2 \gt m_1$ then there is no equilibrium.)

At equilibrium the tension in the string holding up $m_1$ is $T=m_1g$. The tension in the string is transmitted all the way round to $m_2$. The total downward force on the pulley - and therefore also the tension in the spring - is $F=2T=2m_1g$. (We don't need to know the tension in the string below $m_2$.)

The PE stored in the spring is $\frac12kx^2=\frac{F^2}{2k}=\frac{2(m_1g)^2}{k}$.

answered Jan 27, 2017 by sammy gerbil (28,746 points)
selected Jan 28, 2017 by koolman
What mistake I have done ?
The question does not say that the spring is initially relaxed. How do you know that you have reached a position of static equilibrium? I think what you have found is the maximum extension of the spring if $m_1$ is dropped when the spring is relaxed. But then the system will keep oscillating, it will not reach static equilibrium.