Spring Dynamic Kinematics - acceleration

Click For Summary
The discussion focuses on understanding the equation for acceleration in the context of spring dynamics, specifically the expression a = 400 - kx. The poster questions why mass is omitted from the equation, suggesting it should be a = 400 - kx/m, as it seems to imply a relationship to forces. A classmate mentions that mass can be disregarded when it is not relevant to the problem, but fails to clarify this reasoning. It is noted that in this context, k does not represent the spring constant but rather a constant in the linear relationship between acceleration and position. The explanation aims to clarify the role of kx in the acceleration equation without involving mass.
Haizakokaru
Messages
3
Reaction score
0

Homework Statement



2.26_zpsrtzbm2wr.png

Homework Equations


Standard physics equations before forces are introduced into kinematics
U1L6a1.gif


The Attempt at a Solution



I simply looked at Chegg and could not figure out how

a = 400 - kx

when logic dictates that

a = 400 - kx/m

, since this is possibly a sum of forces. Otherwise, it is lost to me how

F= -kx

is suddenly an acceleration quantity. Could someone explain that one line? Everything makes sense save for how

kx

stands alone in an acceleration equation. A classmate of mine who took this class before explained to me that in my dynamics class, mass is dropped from the equation when mass is not relevant to the question. However, he could not describe the reasoning behind it, so I was left somewhat unsatisfied by the answer.[/B]
 
Physics news on Phys.org
In the solution, ##k## does not represent the spring constant. It just represents a constant in the linear relation between ##a## and ##x##.
 
  • Like
Likes Haizakokaru
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
9
Views
4K
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K