# Energy of the system after displacement

A massless pulley has two bodies hanging of masses $$m_1, m_2$$ on both sides. Now $$m_1$$ moves upward by $$h$$ meters and $$m_2$$ moves downward by $$h$$ meters. What is the velocity of the system after the displacement?

I saw the solution that the general equation is $$m_1gh-m_2gh=\frac{1}{2}(m_1+m_2)V^2$$ followed by the explanation that $$m_1gh$$ is the potential energy gained by $$m_1$$ and $$m_2gh$$ is the potential energy lost by $$m_2$$. But when we use formula for potential energy $$mgh$$, isn't $$h$$ the height of the body from ground? If not, then what is $$h$$ which we use in potential energy? Also, could anyone explain the equation above?

• here $m_1gh$ represents change in potential energy of $m_1$ Jun 11, 2021 at 9:14
• Could u pls explain in more detals? Jun 11, 2021 at 9:52

The $$h$$ in $$mgh$$ can be measured from any arbitrary point - the absolute value of the gravitational potential energy doesn't have much meaning, what's important is the $$\textbf{change}$$ in gravitational potential energy between two points in a gravitational field. With reference to the diagram, I can add an arbitrary constant offset $$A$$ to how I label my $$h$$ values, but the change in gravitational potential energy is unchanged as it only depends on the $$\textbf{difference}$$ in $$h$$ values at the two positions.

If I drop an object from rest at height $$h_2$$ and measure its velocity at height $$h_1$$, I'd be using the $$\textbf{Conservation of Energy}$$ equation,

$$\frac{1}{2} m v^2 = mg(h_2-h_1)$$

If I do the same thing, but using a convention where the heights measured are now shifted up by $$A$$, i.e. the object is dropped from height $$h_2+A$$ and its velocity is measured at height $$h_1+A$$, we still get an identical conservation of energy equation. Note that the upper and lower levels still correspond to the same positions in space, I'm just using a different label for their positions. To give a tangible example, in many cases, it's useful to measure the height from the ground level, so implicitly we're saying that $$h=0m$$ occurs at the ground level. Let's say that there's a box whose top surface is 5m above the ground level, so we label its position as $$h=5m$$. The gain in gravitational potential energy when in object is moved from the ground onto the top of the box is then $$mg(5-0) = 5mg$$. If instead we choose to measure the zero reference on the box's surface, i.e. the box's surface corresponds to $$h=0m$$, the ground must correspond to $$h=-5m$$ since it is 5m below the box's surface. The gain in gravitational potential energy when in object is moved from the ground onto the top of the box is then $$mg(0-(-5)) = 5mg$$ and is still the same as before.

The equation $$m_2gh - m_1gh = \frac{1}{2} (m_1 + m_2)v^2$$ is just a conservation of energy equation, the left-hand-side expressing the net loss in gravitational potential energy, which is converted into kinetic energy, expressed on the right-hand-side. If I'm not mistaken I think the masses are swapped, $$m_2gh$$ represents the gravitational potential energy lost by $$m_2$$ as it moved downwards and $$m_1gh$$ represents the gravitation potential energy gained by $$m_1$$ as it moved upwards, so the net amount of gravitational potential energy converted to kinetic energy should be $$m_2gh - m_1gh$$.

💥"But when we use formula for potential energy mgh, isn't h the height of the body from ground?" 🖐️No, h is displacement. If we say the body is at height h above from the ground then potential energy is mgh. But h is displacement in the general case. 💥If not, then what is h which we use in potential energy? 🖐️like i said above it is displacement.I will tell you what I know about potential energy after reading young's unversitu physics. Say there is a mass at a height y1 from ground. It goes down to a height y2 by the action of gravity. Now here the displacement is y1-y2 diwnwards. Force is ng downward. So the angle between force and displacement is zero. Resulting Work W=mg(y1-y2) = - mg(y2-y1) = - (mgy2 - mgy1) = -(final P. E - Initial P. E)

So we define potential energy at height h mgh.that is when y1=h and y2=0. Work done by gravity in bringing from height h to ground is mgh, which is potential energy.

💥Also, could anyone explain the equation above? 🖐️gravitational force is a conservative force.Mechsnical energy is conserved. What is mechanical energy?. Sum of P. E and K. E. Thst initial mechanical energy = final mechanical energy. When m1 moves upwards h displacement , Work is negative mgh.When m2 moves downwards h displacement , Work is positive mgh. By conservation of M.E, (K. E + P. E) initial = (K. E + P. E) final. Lets say initially both m1 and m2 are at a height A above from ground.Both are moving with same velocity V.note that initial K. E is zero because body is rest. So, ( 0+0+m1gA+m2gA) = [1/2 m1 V squared +1/2m2 V squared + m1g(A+h) + m2g(A-h)] Re arranging this you will get m1gh-m2gh = 1/2(m1+m2) V squared 👍hope I answered your question