Potential energy negative, force & displacement co-linear?

AI Thread Summary
The discussion centers on the calculation of gravitational potential energy (V_g) at a specific point where it is found to be negative. The user questions why the gravitational potential energy is negative when both force and displacement are co-linear, which typically results in positive work. The confusion arises from the use of a negative height in the equation V_g = mgh_2, where h_2 is defined as negative due to the chosen reference point. This indicates that the object is below the reference height, leading to a negative potential energy value. The conversation highlights the importance of reference points in determining potential energy values in physics.
CivilSigma
Messages
227
Reaction score
58

Homework Statement


In the following question, in the solution, why is it that when they calculate the gravitational potential energy at point 2 (impact with spring), it is negative? I know that if both the force and displacement act in the same direction, then the work done should be positive. Why is it not the same thing here?
XNuSEne.jpg
 
Physics news on Phys.org
I think they mean both answers could work when the equation is solved...however one answer would obviously make more sense in this case.
 
Oh sorry about that being un-clear. I mean for $$V_g = mgh_2 = (50)(9.81)(-x \sin 20)$$

why is it negative?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top