Block on Incline with oscillation

AI Thread Summary
A block weighing 17.0 N on a 32.0° incline is connected to a spring and needs to be positioned to avoid oscillation. The correct distance down the incline to achieve this is derived from balancing the spring force and gravitational force. The initial calculations using the wrong gravitational component led to incorrect results, but the correct approach involves recognizing that the weight given is already in Newtons. The final solution was reached after correcting the understanding of the forces involved, confirming the importance of accurate force analysis. The discussion highlights the challenges of applying physics concepts correctly in problem-solving.
soccer_09
Messages
8
Reaction score
1

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.
 
Physics news on Phys.org
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top