Block and spring on ramp; find compression

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Homework Help Overview

The problem involves a block on an inclined plane with a spring at the bottom. The block is projected down the ramp towards the spring, and the task is to determine the distance the spring is compressed when the block comes to rest. The context includes concepts from energy conservation, gravitational potential energy, and spring mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles, questioning the definitions and calculations of kinetic energy, gravitational potential energy, and spring work. There is confusion regarding the height variable 'h' and its relationship to the compression of the spring.

Discussion Status

Several participants have offered insights into the energy balance and the relationship between height and spring compression. There is an ongoing exploration of the correct approach, with some participants expressing confusion and others suggesting that assumptions may need to be revisited.

Contextual Notes

Participants note the importance of correctly defining variables and the potential for misunderstanding in the energy balance as the spring compresses. There are references to homework rules that restrict providing direct solutions, emphasizing a focus on understanding the underlying physics.

physics8763
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Member warned to use the formatting template for homework posts.
An inclined plane of angle
θ = 20.0°
has a spring of force constant
k = 525 N/m
fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass
m = 2.31 kg
is placed on the plane at a distance
d = 0.312 m
from the spring. From this position, the block is projected downward toward the spring with speed
v = 0.750 m/s.
By what distance is the spring compressed when the block momentarily comes to rest?

I used equation 1/2mv^2+mgh=1/2kx^2 and got x=.084, which was incorrect.
 
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physics8763 said:
An inclined plane of angle
θ = 20.0°
has a spring of force constant
k = 525 N/m
fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass
m = 2.31 kg
is placed on the plane at a distance
d = 0.312 m
from the spring. From this position, the block is projected downward toward the spring with speed
v = 0.750 m/s.
By what distance is the spring compressed when the block momentarily comes to rest?

I used equation 1/2mv^2+mgh=1/2kx^2 and got x=.084, which was incorrect.
Please show your detailed calculations. And don't omit the HW template. That is the rule at PF.
 
OK, like I said I got x=.084. Basically I have that KE (initial energy) + Gravitational PE is equal to the Work the spring does and solved for x. What did I do wrong?
 
Most likely there is a bug in the homework grading system. .084 is correct.
 
what did you use for 'h' ? You wouldn't know how much potential energy is lost unless you've also solved for the distance by which the spring is compressed
 
physics8763 said:
Most likely there is a bug in the homework grading system. .084 is correct.
Well, did you include the units in your answer?
 
Units are already included and I substituted dsin(theta) for h
 
Maybe there isn't a bug. Hmm I am very confused
 
  • #10
physics8763 said:
An inclined plane of angle
θ = 20.0°
has a spring of force constant
k = 525 N/m
fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass
m = 2.31 kg
is placed on the plane at a distance
d = 0.312 m
from the spring. From this position, the block is projected downward toward the spring with speed
v = 0.750 m/s.
By what distance is the spring compressed when the block momentarily comes to rest?

I used equation 1/2mv^2+mgh=1/2kx^2 and got x=.084, which was incorrect.
untitled-png.90293.png
Hello physics8763. Welcome to PF.

The equation you show could be used to solve the problem, provided that v, h, and, x are properly defined. (True also for m, g, and k)

Physics is much more than an exercise in "hunting for the magic equation".

That equation suggests that you are trying to solve this problem from a "conservation of energy" point of view.As they say "The devil is in the details.", and you have given us very few details regarding how your equation translated into an answer of x=.084 . (Whatever x is.)
 
  • #11
physics8763 said:
Units are already included and I substituted dsin(theta) for h
I don't think h is simply d*sin(theta)
 
  • #12
Does anyone know how to do the problem?
 
  • #13
I need help
 
  • #14
Since all of you don't know how to do it I am going to just look up the answer.
 
  • #15
Hint: h is a function of x
 
  • #16
[Moderator's note: Thread title changed to conform to forum rules. Thread titles must be descriptive of the thread subject matter and identify the type of physics that is involved. Vague titles and pleas for help are not allowed]

Remember that as the spring compresses, the block continues to lose height. What does that do to the energy balance?
 
  • #17
physics8763 said:
Since all of you don't know how to do it I am going to just look up the answer.
I'm sure Steamking, gneill and I all know how to do the problem.

Check what the rules are regarding this Forum. We as HomeWork Helpers must also follow those rules. Putting it bluntly, we do not provide solutions. We are here to help you solve the problem.
 
  • #18
physics8763 said:
Since all of you don't know how to do it I am going to just look up the answer.

Which of these is your desire?
1. The answer.
2. A lesson in how to find the answer.
3, An understanding of the physics involved in reaching the answer.

We can't help with 1. We will help with 2 but only because the goal is 3.
 
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  • #19
Needed quadratic formula: (gsintheta+sqrt(((gsintheta)^2)+(k/m)(v^2+2sintheta*d))/(k/m) gives .1238, the correct answer
 

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