Block on Incline with oscillation

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Homework Help Overview

The problem involves a block on a frictionless incline connected to a spring, where participants are exploring the conditions under which the block would not oscillate when released. The subject area includes mechanics, specifically forces acting on objects on inclined planes and spring dynamics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the block, including gravitational and spring forces. There are attempts to equate these forces to find the position where the block would not oscillate. Some participants express confusion over the correct application of trigonometric functions in their calculations.

Discussion Status

The discussion includes various attempts to solve part (a) of the problem, with some participants sharing their calculations and expressing frustration over incorrect results. One participant eventually claims to have resolved their confusion regarding the weight of the block and its relation to the calculations.

Contextual Notes

Participants mention discrepancies in their calculations and reference other similar problems for comparison. There is a noted lack of responses in previous threads, which may have influenced the flow of discussion.

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Homework Statement



Block on Incline In Fig. 16-35, a block weighing 17.0 N is able to slide without friction on a 32.0° incline. It is connected to the top of the incline by a massless spring of unstretched length 0.475 m and spring constant 110 N/m.

Figure 16-35

(a) How far down the incline would you have to place the block so it would not oscillate when you let go?
1 m
(b) If the block is pulled slightly down the incline from where you placed in part (a) and released, what is the period of the resulting oscillations?



Homework Equations




I got part b, just need part a.

I was told that Force(spring) = Force(gravity)

(Fs) = kx
(Fgrav)= mgcos(theta) since on an incline


The Attempt at a Solution



The force of spring = kx and
The force of grav = mg cos(theta)

By doing this my answer should be x = [ mg cos (theta) ] / k
which comes out to be about 1.284

This showed up as incorrect. Some guidance about what I'm doing wrong is greatly appreciated.

I had posted in another thread a few hours ago and no one responded... but it was one that already had many posts so I guess people just ignored it.
 
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Force of gravity = mg*sin(theta)
 
I already tried that and it was wrong. I got .8026 for my answer and I even tried doing sin and cos of 40 degrees. None of them worked. I looked on cramster and found the same problem since no one had been answering my other post and I found that they got the right answer using the formula I stated above. Their m was 10 N, k was 150 and theta was I believe 37 or 38, I think 38 and the answer was .515 and correct. Having really bad luck I guess. Anything else you might suggest I try? I really can't think of another formula that would work.
 
I figured it out. They give the weight as 17.0 N which is mg so I was doing g squared at first. When I got my answer with that, I added it to the initial distance and got it right. Got it on the 10th try out of 10 as well :) Thanks for trying though hehe. Was just a pure coincidence that the cos worked for the other problem.
 

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