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$\bullet$ A light spring having a force constant of125 $\mathrm{N} / \mathrm{m}$ is used to pull a 9.50 $\mathrm{kg}$ sled on ahorizontal frictionless ice rink. If the sled hasan acceleration of $2.00 \mathrm{m} / \mathrm{s}^{2},$ by how muchdoes the spring stretch if it pulls on the sled (a) horizontally,(b) at $30.0^{\circ}$ above the horizontal?

a) 15.2 $\mathrm{cm}$b) 17.6 $\mathrm{cm}$

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Cornell University

Rutgers, The State University of New Jersey

Hope College

McMaster University

Lectures

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2D kinematics is the study…

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Newton's Laws of Moti…

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When a 2.50 -kg object is …

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A light horizontal spring …

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A force of 800 $\mathrm{N}…

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A $4.00-\mathrm{kg}$ block…

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(a) How much will a spring…

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A 2.50-kg mass is pushed a…

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A 2.50 -kg mass is pushed …

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(II) A 1200-kg car moving …

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One end of a horizontal sp…

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As shown below, two identi…

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question 61 state that a light spring having it forced constant about 25 Newtons. Commuter is pulled is used to police 9.5 kilograms slid on a horizontal friction of service. Ice ring. Actually, this letter is an acceleration of two meters per second squared by how much does this spring stretch if it pulls on the sled horizontally and be, uh, 30 degrees above the horizontal? This is a scenario here for part A pulling horizontally will say the positive X direction, prettify and destruction Republics and exhibit was pulling a spring so spring is attached and pulling it find the amount of extension will call l based on the force being pulled. We want to get with friction, but because I used to be on the right hand side. But because the spring is a restoring force, the spring acts in the opposite direction. So if f is the Applied Force than F s is the spring pours that is the restoring force to counter at that poll in order for this spring slipped to move. That means this restoring first has to be equivalent to the applied for us. We have k l elegant is essentially the spring has to be counteracted by EMI Since we know acceleration, we know mass weaken, simply solve for the extension of the spring T m a over K and we can plug in the numbers here, 9.5 kilograms have an acceleration of two meters per second squared and this is without frictions way. We don't need to consider any friction components to problems on 25 per meter. And so, by solving this, we confined that our final answer is 0.152 meters or, you could say 15.2 center for years RPG, we're now pulling at an angle above the horizontal. In this case, this is our plight force and our springs being pulled in this direction at an angle of 30 degrees above the horizontal. So one thing we can note here is that because we have the same acceleration as we did, the first part of the problem means to be the same forced in the scenario. So with the same force, which means we have the same extension, um el in the X direction. So if again because acceleration, same force, that means we take a look at how much this spring will extend in this scenario, this is TV. If this is the full extension l this is extension the Y direction there for the extension in the X direction at this angle here was 30 degrees, you know, again, based on the previous answer to the extension in the X direction of our spring would be 0.152 meters. Therefore, it needs a better trigonometry relationship between X and L. You know that LX somebody go to the length and are given at a given angle Times co side of 30. Do you agree That's the angle between the two. So that means we can find that the total extension of our spring is actually just LX over co sign 30. We get song for that. We can leave it in centimeters. You're actually doesn't quite matter. But if we have a point 15.2 centimeters the dynamite coastline 30 degrees we would find that are essential. Never spring Here is 17.6 centimeters. There you go.

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