Question

A child s toy consists of a in = 33 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monkey is first hung on the spring and the system reaches equilibrium, the spring has stretched a distance of x = 14.5 cm. as shown in the diagram- This toy is so adorable you pull the monkey down an additional d = 5. 7 cm from equilibrium and release it from rest. and smile with delight as it bounces playfully up and down. Assume that the system has zero gravitational potential energy at the lowest point of the motion. Derive an expression for the total mechanical energy, Eequilibrium, of the system as the monkey passes through the equilibrium position in terms of m, x, d, g, k, and the speed of the monkey, ve.

Answer 1

Answer:

$E = \frac{1}{2}kx^2 + \frac{1}{2}mv_e^2 + mgd$

Explanation:

As we know that the spring constant of the spring is k

so at equilibrium position we will have

$mg = kx$

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

$E = \frac{1}{2}kx^2 + \frac{1}{2}mv_e^2 + mgd$

here we know that

$v_e$ = speed of the monkey

m = mass of the monkey

k = spring constant