Answer:
[tex]E = \frac{1}{2}kx^2 + \frac{1}{2}mv_e^2 + mgd[/tex]
Explanation:
As we know that the spring constant of the spring is k
so at equilibrium position we will have
[tex]mg = kx[/tex]
Now when we push the monkey downwards by a distance "d"
and taking this as our gravitational reference where potential energy is assumed to be zero
so we will have
Total mechanical energy at equilibrium is given as
[tex]E = \frac{1}{2}kx^2 + \frac{1}{2}mv_e^2 + mgd[/tex]
here we know that
[tex]v_e[/tex] = speed of the monkey
m = mass of the monkey
k = spring constant
At equilibrium position and using the terms of m, x, d, g, k, and the speed of the monkey, ve. we will have an expression for total mechanical energy as
[tex]E = \frac{1}{2}kx^2 + \frac{1}{2}mv_e^2 + mgd[/tex]
Question Parameters:
Generally the equation at equilibrium position is mathematically given as
mg = kx
Hence,Total mechanical energy is give as
[tex]E = \frac{1}{2}kx^2 + \frac{1}{2}mv_e^2 + mgd[/tex]
Therefore we have that at equilibrium position and using the terms of m, x, d, g, k, and the speed of the monkey, ve. we will have an expression for total mechanical energy as
[tex]E = \frac{1}{2}kx^2 + \frac{1}{2}mv_e^2 + mgd[/tex]
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