Calculating Spring Stretch on a Ramp with Varying Acceleration and Friction

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The discussion revolves around calculating the stretch of a spring attached to a block on a ramp under various conditions. For part a, the force on the spring is derived from the gravitational force component acting parallel to the ramp, resulting in a stretch of approximately 0.196 meters. In part b, the challenge is to incorporate the block's acceleration of 4 m/s² into the calculations, suggesting the need to analyze the forces using a free body diagram (FBD). The participants clarify the relationship between gravitational force, spring force, and acceleration, emphasizing the importance of understanding the net forces involved. Overall, the thread focuses on applying physics principles to solve for spring stretch under different scenarios.
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1. Homework Statement
We are dealing with a block of 20 kg mass on a ramp that is at an angle of 30 degrees. The spring constant given is 500 N/m.

Picture- https://flic.kr/p/A3aCsM
The only difference is that that we are supposed to be holding the spring stationary, it isn't attached to a wall.

Assume frictionless for a and b.The questions wants to know a) how far the spring will stretch at rest

b)You pull so the block is accelerating at 4m/s^2 up the ramp. How far does the spring stretch?

c)Now you are to assume you are on a ramp with the following coefficients of friction us=0.3 and uk= 0.2. If you are initially holding the free end of the spring and you pull the spring so that the block is accelerating at 4m/s up the ramp, how far is the spring stretched?

Homework Equations


Fspring=kx

The Attempt at a Solution



a) I used F=mgsinΘ to find that the force is 98 N, then I used F=kx to get x=F/k= 98N/500N/m=0.196m
I think this is how to do it, but I was wondering if someone could help me understand where F=mgsinΘ came from? Is the force of the spring equal to the x component of the weight because the block is attached at rest?

b)
I am at a full loss of how to start the second part. I think the question means that since we are holding the free end of the spring, we just pull it back. But I'm not sure how acceleration plays into any of the equations for the spring unless since.. Fspring=mgsinΘ=kx and ∑Fx=mgsinΘ=ma because we pull on it, we can set ma=kx, and solve for x? That's the only thing I can think to do.
 
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For a) In order to determine the force on the spring, you have to determine how much of the vertical gravity force is directly opposite to the direction that you are pulling the string. To help yourself understand first, put a dot in the center of the block and second, draw a short vertical line downward (this is the direction of the force of gravity), next, from the dot, draw another line parallel to the ramp going down the ramp; and, finally starting at the bottom of your first vertical line, draw a line upward perpendicular to the slope of the ramp until it intersects your line going parallel with the ramp. By now you will have formed a 30 degree right triangle with it hypotenuse being the direction of the force of gravity and the short line being the amount of the gravity force that is directly away from your spring and the length of that short line is = F gravity x sin 30 degrees. Additionally, the other leg of the triangle represents the amount of gravity that is acting to create the friction of the block and it = F gravity x cos 30 degrees.

Does that help?
 
Yes, that helps clear up the reason behind part A a lot, thank you! :smile:

For part b, was I totally off on how to figure out the stretch of the spring if the acceleration is 4 m/s^2?
 
Tim Wellens said:
For part b, was I totally off on how to figure out the stretch of the spring if the acceleration is 4 m/s^2?
Just draw the usual FBD. What are the forces, what is the net force, etc?
 
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