1
$\begingroup$

Now Work-Energy Theorem is derived by considering the change in Kinetic Energy between 2 points with position vectors $\vec{r}$ and $\vec{r}+\vec{dr}$.

Now shouldn't the Kinetic Energy be differentiated with respect to displacement $\vec{dr}$?

Here we have differentiated with respect to $dt$.

So the change in Kinetic Energy represented here is change in Kinetic Energy in a small interval of time $dt$.

$\int{dT}$= $\int{\vec{F}.\vec{dr}}$

Now the Change in Total Energy is equal to Change in Kinetic Energy + Change in Potential Energy.

Now the term Change in Kinetic Energy $∆{T}$ is change observed in Kinetic Energy in a small time $dt$ ,but the change in Change in Potential Energy $∆U$is the change observed in Potential Energy due to a small displacement $dr$.

Now this seems wrong ,shouldn't both the quantities supposed to calculated with respect to same quantity while evaluating the total change in Mechanical Energy?

the Total change in Mechanical Energy ( change observed with respect to time / displacement ? ) = Change in Kinetic Energy $∆T$ ( change observed in Kinetic Energy in a small time) + Change in Potential Energy $∆U$(the change observed in Potential Energy due to a small displacement).

Change in Mechanical Energy is supposed to be zero with respect to time / displacement?

https://i.stack.imgur.com/Vi5Zn.jpg

$\endgroup$
1

2 Answers 2

0
$\begingroup$

Forget about potential energy in the derivation of the work-energy theorem. The theorem states that change in kinetic energy $\Delta K$ equals total work done on an object $W$. Work done by the gravitational force is defined as $W_g = -\Delta U_g$ where $U_g = mgy$ is gravitational potential energy (valid only close to planet surface!) and positive $y$ direction points upwards. Note that work $W_g$ is already considered to be part of total work $W$ in the work-energy theorem.

The work by definition is

$$dW = F \cdot dx$$

where $F$ is net force on the body that acts in the same direction as displacement $dx$.

The kinetic energy abstraction can be derived as follows

$$dW = m a \cdot dx = m \frac{dv}{dt} \cdot dx = m \frac{dx}{dt} \cdot dv = m v \cdot dv$$

Now take integral of the above equation

$$\int_{W_i}^{W_f} dW = \int_{v_i}^{v_f} mv \cdot dv$$

which gives final form of the work-energy theorem

$$W_f - W_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$$

Kinetic energy is $K = \frac{1}{2} m v^2$ and the above equations becomes

$$\boxed{\Delta K = W}$$

where $W = W_f - W_i$ is the total work done on the object.

Change in Mechanical Energy is supposed to be zero with respect to time / displacement ?

This is true only if there are no non-conservative forces acting in the system, such as friction force. Otherwise, the mechanical energy changes with time. Just study simple damped oscillators.

$\endgroup$
0
$\begingroup$

Note that the key phrase in the statement of the work energy theorem shown in your image is the "net force" over the path, which is equivalent to the "net work" done over the path.

In applying the work energy theorem, the forces over the path include any conservative forces (gravity, electromagnetic, elastic) associated with changes in potential energy over the path well as any dissipative forces such as friction. The term "net" is critical because these forces can be either positive or negative doing positive or negative work. When a force does positive work on an object it transfers energy to the object. When a force does negative work on an object it takes energy away from the object. The net work done on the object is the sum of the two and equals the change in kinetic energy.

When there is an increase in potential energy of the system it is because a conservative force has done negative work. When there is a decrease in potential energy it is because a conservative force has done positive work. Consider the following three scenarios which assume no air resistance.

Scenario 1:

Let the two end points of the path be point 1 on the ground and point 2 a height $h$ above the ground where $g$ is constant over the path. I lift an object of mass $m$ initially at rest at point 1 and bring it to rest at point 2. I do positive work of $+mgh$ and gravity does an equal amount of negative work $-mgh$ (because the force of gravity is opposite to the displacement of the object) for a net work of zero and change in kinetic energy of zero, per the work energy theorem. Gravity took the energy I gave the object and stores it as gravitational potential energy (GPE) of the earth-object system.

Scenario 2:

I now release the object from point 2. Gravity now does positive work of $+mgh$ giving the object kinetic energy of $\frac{1}{2}mv^{2}=mgh$ prior to impact at point 1. Since now only gravity does work, the net work done on the object is the work done by gravity and equals the change in kinetic energy.

Scenario 3:

The path is now on a horizontal surface with friction. An object with having kinetic energy $\frac{1}{2}mv^2$ is released at point 1 and comes to rest at point 2 a distance $d$ from point 1. The only force acting on the object is kinetic friction and equals $-\mu_{k}mg$. The force and work done by friction is negative as it is opposite to the displacement of the object. The net work done is thus the work done by friction so $W_{net}=-\mu_{k}mgd$. Per the work energy theorem this equals the change in kinetic energy or $-\frac{1}{2}mv^2=-\mu_{k}mgd$.

Change in Mechanical Energy is supposed to be zero with respect to time / displacement?

Assuming you mean total mechanical energy, then it will not change (i.e. $\Delta KE+\Delta U =0$) (will be conserved) with respect to time and/or displacement, under the following conditions: (1) There are no dissipative forces (e.g., friction) doing work and (2) The defined system is "isolated', i.e., there are no external forces doing work on the system. Condition 2 requires that we define the system.

Let the earth-object be the system for scenarios 1 and 2 (because potential energy is a system property) and the object be the system for scenario 3.

In scenario 1 mechanical energy is not conserved because condition 2 is not met. There is an external force (me) acting on the system.

In scenario 2 mechanical energy is conserved ($\Delta KE+\Delta U=0$) because both conditions are met (no external force acting on system and no dissipative force such as friction).

In scenario 3 mechanical energy is not conserved because condition 1 is not met.

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.