Work-Energy Theorem - A conceptual doubt [closed]

Question

I lately have been confused by the work energy theorem which states that:

Work done by all the forces on an object is equal to the change in kinetic energy.

However, I have a doubt

Consider we have an object resting on the ground. We lift the object from the group and bring it to a height say $h$. We ensure that the work done in bringing the object above the ground is done without giving it any velocity (By applying a force $F = mg + dF$, where $dF > 0$). Now, the work done by gravity = -mgh. (h<<R). Now, applying Work-energy theorem:

δ(K.E) = -mgh (neglecting air friction)

But in this case , change in Kinetic energy will be zero... hence implying that mgh = 0, and h = 0 but this is not the case.

So could anyone please tell me what am I doing wrong here?

Answer 1

The problem is that you're not including the work done by your hand. If you include this, then the net work is $$W_{\text{net}} = -mgh + W_{\text{hand}}$$ If the pen starts and ends at rest, then the change in kinetic energy is zero, regardless of the speed you raise the pen. So the work-energy theorem says $$-mgh + W_{\text{hand}} = \Delta K = 0$$ and we get $$W_{\text{hand}} = mgh$$ The interpretation of this is that we've done work on the Earth-pen system with our hand to increase the potential energy of the system by $mgh$.

Answer 2

While applying the WORK-ENERGY THEOREM you have to make sure that you are accounting respective work done by all the forces acting on the body.

In your case, you have omitted the work done by the force ($F$) - applied by you.

So correct expression would be:

$$W_{mg}+W_{F}=\Delta KE$$ $$-mgh + F.h = \Delta KE = 0$$

Since $F$ is nearly equal to $mg$, so putting $F=mg$ in above equations proves the equality.

Answer 3

The sum of normal force from your hand and gravity are doing work on the pen in this example, when you displace it. If the normal is greater than gravity, then net work is done on the pen, as you rise it.

By the theorem stated, this gives the pen kinetic energy because while it is rising, it has some velocity. If the velocity was high enough, the pen would continue flying when you stop your hand (if you were to try this outside, hypothetically, put a pen on your open palm and raise your hand pretty fast).

In your example, when you stop moving your hand, any remaining velocity the pen has (which is tiny) will still propel it up and gravity will do work on the pen which means it will lose the kinetic energy and hence go to $0$.

In summary, at any point when the normal force is greater than gravity, there is a net force on the pen and thus displacing it will result work and so by the theorem, a in a gain of kinetic energy. This need not be the case for the entire rising process ( this easier to picture with elevator example linked).

Non-energy line of logic for comparison: at any point the normal force is greater than gravity, there is a net force on the pen and thus it accelerates by Newton's Second Law. This means it gains velocity.

Related question: Is it possible for the Normal Force to do work?

Answer 4

The work done by all forces is $W = h \big( -mg + mg + dF \big) = h dF$. Your calculation misses the work done by the applied force. So the change in kinetic energy is infinitesimal. Also your process takes infinitely long.

You can do the same with finite intermediate speeds, just take an arbitrary function for the height in dependence of time $h(t)$, then you get the sum of all forces from Newton's law $F = ma$: $F(t) = m \ddot h(t)$.

The force you have to supply externally is consequently $m \ddot h(t) + mg$. (Assuming the force of gravity acts as $F_G = -mg$).

The work done by all forces is then, again, the difference of the kinetic energy (as stated) $$m \int_0^t dt\, \dot h(t) \ddot h(t) = m \int_0^t \frac 1 2 \big(\partial_t \dot h(t)\big)^2 = \left. \frac 1 2 m \big( \dot h(t) \big)^2 \right|_0^t = T(t) - T(0).$$