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Chapter 6: Problem 44

A block of mass \(0.773 \mathrm{~kg}\) on a spring with spring constant \(239.5 \mathrm{~N} / \mathrm{m}\) oscillates vertically with amplitude \(0.551 \mathrm{~m}\). What is the speed of this block at a distance of \(0.331 \mathrm{~m}\) from the equilibrium position?

Short Answer

Expert verified
Answer: The speed of the block at a distance of 0.331 meters from the equilibrium position is 7.334 m/s.

Step by step solution

01

Calculate the initial total mechanical energy

Initially, when the spring is fully compressed or stretched, the block is at the amplitude (\(A = 0.551 m\)) and has maximum potential energy and zero kinetic energy. The potential energy of the spring can be found using Hooke's Law: \(PE = \frac{1}{2} kA^2\), where \(k\) is the spring constant and \(A\) is the amplitude. Given: \(k = 239.5 \mathrm{~N}/\mathrm{m}\) \(A = 0.551 \mathrm{~m}\) \(PE = \frac{1}{2} (239.5 \mathrm{~N}/\mathrm{m}) (0.551 \mathrm{~m})^2 = 36.398 \mathrm{~J}\) At the amplitude, the initial total mechanical energy is only potential energy: \(E_{initial} = PE = 36.398 \mathrm{~J}\)
02

Calculate total mechanical energy at the given distance

Let's call \(x\) the distance from the equilibrium position at which we want to find the speed of the block. Given: \(x = 0.331 \mathrm{~m}\) At this point, the total mechanical energy is the sum of potential energy \(PE_x\) and kinetic energy \(KE_x\): \(E_x = PE_x + KE_x\) Since the energy is conserved, we can say that: \(E_{initial} = E_x\)
03

Calculate the potential energy at the given distance

The potential energy at distance \(x\) can be found using Hooke's Law: \(PE_x = \frac{1}{2} kx^2\) \(PE_x = \frac{1}{2} (239.5 \mathrm{~N}/\mathrm{m}) (0.331 \mathrm{~m})^2 = 13.078 \mathrm{~J}\)
04

Find the kinetic energy at the given distance

From the conservation of energy equation (\(E_{initial} = E_x\)), we can find the kinetic energy at distance \(x\): \(KE_x = E_{initial} - PE_x\) \(KE_x = 36.398 \mathrm{~J} - 13.078 \mathrm{~J} = 23.320 \mathrm{~J}\)
05

Calculate the speed of the block at the given distance

We can use the kinetic energy formula to find the speed of the block: \(KE_x = \frac{1}{2} mv^2\), where \(m\) is the mass of the block and \(v\) is the speed. \(v = \sqrt{\frac{2 KE_x}{m}}\) Given: \(m = 0.773 \mathrm{~kg}\) \(v = \sqrt{\frac{2 (23.320 \mathrm{~J})}{0.773 \mathrm{~kg}}} = 7.334 \mathrm{~m/s}\) The speed of the block at a distance of \(0.331 \mathrm{~m}\) from the equilibrium position is \(7.334 \mathrm{~m/s}\).

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