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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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Most popular questions from this chapter

You use your hand to stretch a spring to a displacement \(x\) from its equilibrium position and then slowly bring it back to that position. Which is true? a) The spring's \(\Delta U\) is positive. b) The spring's \(\Delta U\) is negative. c) The hand's \(\Delta U\) is positive. d) The hand's \(\Delta U\) is negative. e) None of the above statements is true.

A runner reaches the top of a hill with a speed of \(6.50 \mathrm{~m} / \mathrm{s}\) He descends \(50.0 \mathrm{~m}\) and then ascends \(28.0 \mathrm{~m}\) to the top of the next hill. His speed is now \(4.50 \mathrm{~m} / \mathrm{s}\). The runner has a mass of \(83.0 \mathrm{~kg} .\) The total distance that the runner covers is \(400 . \mathrm{m}\) and there is a constant resistance to motion of \(9.00 \mathrm{~N}\). Use energy considerations to find the work done by the runner over the total distance.

A child throws three identical marbles from the same height above the ground so that they land on the flat roof of a building. The marbles are launched with the same initial speed. The first marble, marble \(\mathrm{A}\), is thrown at an angle of \(75^{\circ}\) above horizontal, while marbles \(\mathrm{B}\) and \(\mathrm{C}\) are thrown with launch angles of \(60^{\circ}\) and \(45^{\circ}\), respectively. Neglecting air resistance, rank the marbles according to the speeds with which they hit the roof. a) \(A

A mass of \(1.00 \mathrm{~kg}\) attached to a spring with a spring constant of \(100 .\) N/m oscillates horizontally on a smooth frictionless table with an amplitude of \(0.500 \mathrm{~m} .\) When the mass is \(0.250 \mathrm{~m}\) away from equilibrium, determine: a) its total mechanical energy; b) the system's potential energy and the mass's kinetic energy; c) the mass's kinetic energy when it is at the equilibrium point. d) Suppose there was friction between the mass and the table so that the amplitude was cut in half after some time. By what factor has the mass's maximum kinetic energy changed? e) By what factor has the maximum potential energy changed?

A 80.0 -kg fireman slides down a 3.00 -m pole by applying a frictional force of \(400 .\) N against the pole with his hands. If he slides from rest, how fast is he moving once he reaches the ground?
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