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Chapter 6: Problem 53

The length of a spring increases by \(7.2 \mathrm{cm}\) from its relaxed length when a mass of \(1.4 \mathrm{kg}\) is hanging in equilibrium from the spring. (a) What is the spring constant? (b) How much elastic potential energy is stored in the spring? (c) A different mass is suspended and the spring length increases by \(12.2 \mathrm{cm}\) from its relaxed length to its new equilibrium position. What is the second mass?

Short Answer

Expert verified
Answer: 0.4959 J

Step by step solution

01

(a) Determine the force applied on the spring by the hanging mass

To calculate the constant spring, we first need to find the force applied on the spring by the mass hanging in equilibrium. Using the formula \(F = mg\), where \(F\) is the force, \(m\) is the mass, and \(g\) is the acceleration due to gravity (\(9.81 m/s^2\)), we can find the force applied: \(F = 1.4 * 9.81 = 13.734 N\)
02

(a) Calculate the spring constant using Hooke's law

Hooke's law states that the force applied on a spring is proportional to the displacement of the spring, or \(F = k\Delta{L}\), where \(F\) is the force, \(k\) is the spring constant, and \(\Delta{L}\) is the change in length of the spring. Rearranging the formula for \(k\), we get \(k = \frac{F}{\Delta{L}}\). Plug in the values: \(k = \frac{13.734 N}{0.072 m} = 190.75 N/m\)
03

(b) Calculate the elastic potential energy stored in the spring

Elastic potential energy stored in a spring is given by the formula \(PE = \frac{1}{2} k (\Delta{L})^2\). Using the spring constant obtained in part (a) and the change in length, we can find the elastic potential energy: \(PE = \frac{1}{2} (190.75 N/m) (0.072 m)^2 = 0.4959 J\)
04

(c) Determine the force applied on the spring by the second mass with the new spring length

The new equilibrium position increased the spring length by \(0.122 m\), we can use Hooke's law \(F = k\Delta{L}\) again to determine the force: \(F = (190.75 N/m) * (0.122 m) = 23.2515 N\)
05

(c) Calculate the second mass using the force obtained from the new spring length

Now that we have the force applied by the second mass, we can use the equation \(F = mg\) to find the mass. Rearrange the equation for \(m\): \(m = \frac{F}{g} = \frac{23.2515 N}{9.81 m/s^2} = 2.369 kg\) The spring constant is \(190.75 N/m\), the elastic potential energy stored in the spring is \(0.4959 J\), and the second mass is \(2.369 kg\).

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