Kevin S. Huang

2007: Problem 19

Topic: Energy
Concepts: Conservation of energy, Spring potential energy

Solution:

We have a non-Hookian spring with force F(x)=−kx2F(x)=-kx^2 so its corresponding potential energy is

U(x)=−∫0xF(s)ds=∫0xks2dx=13kx3U(x)=-\int_0^x F(s)\,ds=\int_0^x ks^2\,dx=\frac{1}{3}kx^3

Once the mass is released, it will momentarily be at rest when the loss in gravitational potential energy balances the gain in spring potential energy:

mgh=13kd3mgh=\frac{1}{3}kd^3

Since θ=30∘\theta=30^{\circ} and h=dsin⁡θ=d/2h=d\sin\theta=d/2,

mgd2=kd33\frac{mgd}{2}=\frac{kd^3}{3}
d=3mg2kd=\sqrt{\frac{3mg}{2k}}

so the answer is A.

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Response

  1. clearmentality612568d046 Avatar
    clearmentality612568d046

    Hi Kevin, I tried using a different approach for this problem where I drew the forces acting on the block and equated the spring force to the mgsin(30) component of gravity that is parallel to the slope. However, when using this method I don’t get any of the answers that are listed, including the correct one. I was wondering why this method doesn’t work, and if it does, where I went wrong.

    Reply

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