Effective spring constant of two springs latched headon

AI Thread Summary
To calculate the effective spring constant of two springs joined side by side, the correct approach involves understanding how the forces and displacements relate to each spring's constants, k1 and k2. The initial equations provided, F = -k1/x1 and F = -k2/x2, may not accurately represent the system without further context. The discussion suggests reviewing the subsequent lines in the derivation for clarity. Additionally, a Wikipedia article on series and parallel springs can provide further insights into the topic. Understanding the configuration of the springs is crucial for determining the effective spring constant correctly.
duckblase
Messages
1
Reaction score
0
In the Princeton Review for AP Physics, there is a question on calculating the effective spring constant when two springs are joined side by side, like so:

spring 1 spring 2
|~~~~~~ ~~~~~~[mass]

Their spring constants are k1 and k2.

How would you go about answering the question? The book starts by saying F = -k1/x1 and F = -k2/x2, where F is the net force, and x1 and x2 are the displacements for each spring, but this does not seem right to me. Is it?

Thanks!
 
Physics news on Phys.org
Hi, and welcome to Physics Forums! :smile:

It's probably not right, but what are the next couple of lines in the derivation they are showing? Once we know that, we can see what they are angling at. :wink:
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top