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$\bullet$ $\bullet$ Inside a NASA test vehicle, a 3.50 -kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The force constant of the spring is 225 $\mathrm{N} / \mathrm{m} .$ The vehicle has a steady acceleration of $5.00 \mathrm{m} / \mathrm{s}^{2},$ and the ball is not oscillating. Suddenly, when the vehicle's speed has reached $45.0 \mathrm{m} / \mathrm{s},$ its engines turn off, thus eliminating its acceleration but not its velocity. Find (a) the amplitude and (b) the frequency of theresulting oscillations of the ball. (c) What will be the ball's maximum speed relative to the vehicle?

a) 0.0778 $\mathrm{m}$b) 1.28 $\mathrm{Hz}$c) 0.624 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 11

Elasticity and Periodic Motion

Equilibrium and Elasticity

Periodic Motion

Cornell University

Simon Fraser University

McMaster University

Lectures

04:12

In physics, potential ener…

02:18

In physics, an oscillation…

08:11

A $2.00-\mathrm{kg}$ objec…

07:28

A 2.00 -kg object is attac…

04:28

$\bullet$ A 0.500 kg glide…

01:53

A 175-g glider on a horizo…

02:01

A $2.00-\mathrm{kg}$ block…

03:25

A $100 \mathrm{g}$ ball at…

03:21

A $175-$ glider on a horiz…

03:51

A 175 -g glider ona horizo…

03:26

A 1.50-kg ball and a 2.00 …

13:37

In a spring gun, a spring …

03:39

A 2.00 -kg object on a fri…

04:16

Suppose you attach the obj…

04:40

An object of mass $m=$ 5.0…

06:42

$\bullet$ A 0.150 $\mathrm…

09:05

$*$ A spring with a spring…

05:27

A 1.00 -kg glider attached…

An automobile can be consi…

06:48

A $1.00-\mathrm{kg}$ glide…

05:19

An object of mass 5.00 $\m…

05:26

A 0.150 -kg toy is undergo…

during this problem, we're going to work in the frame of reference of the car. Once the engines shut off, the acceleration is equal to five years for something, and this will determine the amplitude of the motion. You know where it's actually determine the value, the amplitude We're going to use Newton's second law, which tells us that the force it's secret to the mass times acceleration but hooks a lot tells us that the forces negative K X. It's the way of that. Solving for X gives X is master's acceleration over K. I'm dropping the native sign for the moment, and a maximum X is equal to a. And so if we plug in the acceleration here, we can figure out the X value attained at its maximum, and that will be called the amplitude. And so where were we plug in the values here we get. The A is equal to 0.778 meters, and that's the answer to part it for Part B. We want to figure out what the frequency of the motion is, and so we can use the Formula One over to Pie Times, the square of K over him is equal to one over to him. And now I'm going to plug in values they give us so to 25 for the spring constant and the masses 3.5 kilograms. This turns out to be 1.28 and the units on frequents your herds. So for part C, we had to figure out what the maximum velocity is once it starts this periodic motion. And so a peak philosophy a peak speeds you is equal to zero, the potential is equal zero. And so conservation of energy gives us that 1/2 k squared, which is the energy is equal to only the kinetic energy. When you're a zero and the connectors use 1/2 em and I'm gonna label V by the max since one new is equal to zero V is equal to the max. I could solve this for me, Max and I get the square root of K around time's amplitude, and I can plug in everything on there and side here himself. To get 0.6. It's a six there, 0.6 6 6 to 4 meters per second, and that's a maximum velocity. And that's the problem.

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