Qns on potential energy and forces

AI Thread Summary
The potential energy of a body at point P, given by V = kx^2, leads to a force of 2kx directed towards the reference point O. The initial misunderstanding involved calculating force as kx based on the work-energy principle, but this applies only when force is constant. In cases of variable force, the correct approach is to differentiate the work done with respect to distance, resulting in the force being the negative derivative of potential energy. This indicates that the force acts towards the origin, confirming the attraction towards point O. Understanding the relationship between potential energy and force is crucial for accurately determining the dynamics of the system.
coffeebeans
Messages
7
Reaction score
0
Hello, i hope i posted this in the correct section..

I have a question:
the potential energy of a body when it is at point P a distance x from a reference point O is given by V = kx^2, where k is a constant. what is the force acting on the body when it is at P?

the correct ans is 2kx in the direction of PO

the ans i got is kx in the direction OP. as work = force x distance, so force required to bring the body to point p is kx^2 / x .

could some please explain why is the correct ans so? thnk u so much!
 
Last edited:
Physics news on Phys.org
Work is "force times distance" as long as the force is constant- if the force is a variable then the work is given by \int f(x)dx. Notice that this involves a "constant of integration". It is standard to choose that constant so that potential energy is 0 at some specific point; potential energy is always relative to some given point.

Going the other way, if work is a constant times distance, W= Cx then force is that constant: F= Cx/x= C. But if a more general function then F= \frac{dW}{dx}.
 
The potential energy is positive. So there is an attraction toward the origin.

The force = - (dW/dx) = -2kx.
So the force is directed from the point P to the point O, the origin.
bye
 
Thnx alot!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top