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Chapter 10: Problem 44

A \(0.50-\mathrm{kg}\) object, suspended from an ideal spring of spring constant \(25 \mathrm{N} / \mathrm{m},\) is oscillating vertically. How much change of kinetic energy occurs while the object moves from the equilibrium position to a point \(5.0 \mathrm{cm}\) lower?

Short Answer

Expert verified
Answer: The change in kinetic energy of the object is 0.03125 J.

Step by step solution

01

Calculate the elastic potential energy at equilibrium

At the equilibrium position, the elastic potential energy is zero, since the spring is neither compressed nor stretched.
02

Calculate the elastic potential energy at the position 5 cm lower

To calculate the elastic potential energy when the object is 5 cm lower, we need to find the amount the spring is stretched. The displacement from the equilibrium position is 5 cm, which we convert to meters: \(x = 5.0 \mathrm{cm} \times \frac{1 \mathrm{m}}{100 \mathrm{cm}} = 0.05 \mathrm{m}\). Now, using Hooke's Law, the elastic potential energy (PE_elastic) is given by: \(PE_{elastic} = \frac{1}{2} kx^2\) where \(k\) is the spring constant and \(x\) is the displacement. Plugging in the given values, we have: \(PE_{elastic} = \frac{1}{2} (25 \mathrm{N/m}) (0.05 \mathrm{m})^2\) \(PE_{elastic} = 0.03125 \mathrm{J}\)
03

Finding the change in kinetic energy

Since we know that the total mechanical energy is conserved, the change in elastic potential energy should be equal to the change in kinetic energy. So, the change in kinetic energy (ΔKE) is given by: \(\Delta KE = PE_{elastic} - 0\) \(\Delta KE = 0.03125 \mathrm{J}\) The change in the kinetic energy of the object while moving from the equilibrium position to a point 5 cm lower is \(0.03125 \mathrm{J}\).

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