What type of of force is applied force? Conservative ,Nonconservative

[Miraj Kayastha]

In a closed loop when we apply an applied force on an object the object starts at point A and stops at point A.

Since the displacement is 0, Work done by the applied force on the object is = F x s x cosθ
= 0

So the net work done by the applied force is 0 but why is applied force a non-conservative force?

 

[ehild]

When the force is not constant during the motion the work is calculated by integration. Have you studied calculus?

ehild

 

[Miraj Kayastha]

Yes I have.

 

[ehild]

The work done along a curved path is $W=\int F_s d_s$ where Fs is the component of the force tangent to the segment of the curve and ds is the length of the line segment. If you apply a force of constant magnitude, always tangent to the curve, $W=F \int d_s =F L$ where L is the length of the curve. The work is not zero along a closed loop.

If you push a crate along a rough horizontal surface with force F you do F*D work during D distance. When you push it back with force of the same magnitude, you do F*D work again. The displacement is zero, but the net work is not.

In general, the work depends on the way it is done. The conservative forces are exceptions, their work does not depend on the path taken between the initial and final points.

ehild

 

[Miraj Kayastha]

If you push a crate along a rough horizontal surface with force F you do F*D work during D distance. When you push it back with force of the same magnitude, you do F*D work again. The displacement is zero, but the net work is not.
ehild

Here when you said "when we push it back with the same magnitude, you do F x D work again" I think we should also account the direction of the force because the force is constant both in magnitude and direction throughout the motion. So shouldn't the work done by the force on the crate when crate is coming back be - F x D ?

And then the net work is zero.

 

[jbriggs444]

Here when you said "when we push it back with the same magnitude, you do F x D work again" I think we should also account the direction of the force because the force is constant both in magnitude and direction throughout the motion. So shouldn't the work done by the force on the crate when crate is coming back be - F x D ?

The D in the work formula is not a "distance". It is a "displacement". The distinction is that a distance is always positive and has no direction. A displacement has a direction.

When you push the crate back toward the starting point, both force and direction are reversed. So it's -F x -D.

 

[ehild]

The elementary work can be written either as dot product of the force vector with the displacement vector, $dW=\vec F \cdot \vec{dr}$ or as the product of two scalars, Fs the component of force along the path taken and the elementary path length, ds dW=F_sd_s.

ehild