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

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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
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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.
 
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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 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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