How to Calculate Variable Force on a Particle with Varying Coordinates?

AI Thread Summary
To calculate the work done by a variable force on a particle, integration is necessary due to the changing nature of the force. The provided formula, W = -k(y i + x j)d, applies only to constant forces, which is not suitable in this case. Accurate coordinates of the particle are essential for determining the correct value of d. The discussion emphasizes the importance of integrating the force over the path to obtain the work done. Understanding these concepts is crucial for solving problems involving variable forces.
rudransh verma
Gold Member
Messages
1,067
Reaction score
96
Homework Statement
A force ##\vec F=-k(y\hat i+ x\hat j)## where k is constant acts on the particle moving in x-y plane from origin, the particle is taken along the positive x axis to the point (a,0) and parallel to y axis to a point (a,a). The total W by the Force on the particle is
Relevant Equations
W=F.d
##W=-k(y\hat i+ x\hat j)d##. I am not getting the coordinate of particle correctly so that I can find the value of d. Also the force is varying.
 
Physics news on Phys.org
Since the force is varying, you'll need to integrate. The formula you gave is for constant force. And the coordinates are given.
 
Doc Al said:
Since the force is varying, you'll need to integrate. The formula you gave is for constant force. And the coordinates are given.
Oh yes! Thanks.
 
Thread 'Voltmeter readings for this circuit with switches'

Thread 'Voltmeter readings for this circuit with switches'

TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
View full post »
Thread 'Correct statement about a reservoir with an outlet pipe'

Thread 'Correct statement about a reservoir with an outlet pipe'

The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
View full post »

Similar threads

Where on a rigid body is a resultant force applied?
Replies
11
Views
1K
What is the induced magnetic force on this closed current loop?
Replies
12
Views
2K
How to obtain correct negative sign in calculation of motional emf?
Replies
2
Views
935
Vector is a function of its position or not?
Replies
16
Views
1K
Bead sliding along wire with constant horizontal velocity -- Shape of wire?
Replies
2
Views
1K
Motion of a particle under a tangential force
Replies
3
Views
1K
Effect of different magnetic fields on a magnetic dipole
Replies
25
Views
1K
Modeling a mass falling from one side of the Earth to the other
Replies
1
Views
973
Calculation of torque and force on magnetic dipole near infinite wire
Replies
3
Views
1K
Fundamental Forces: Force of one charged particle on another
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top