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3 years ago

What is the most important thing to remember about how to calculate Mechanical Energy?

Physics

1 answer:

nataly862011 [7]3 years ago

6 0

Answer:

The mechanical energy of a classical system is simply equal to the sum of its kinetic and potential energies.

E=T+V

Where the kinetic energy T is given by the equation

T=12mv2

and the potential energy V differs depending on whatever forces are acting on the system.

If the force acting on the object is gravity (a common case) the potential energy is given by

V(r)=GMmr

where the Ms represent the mass of the two interacting objects (The system and the earth, for example), r is the distance between their centers of mass, and G is the universal gravitational constant (=6.67e-11 N m^2 kg^-2)

In simple, close to earth, situations, it takes the simpler form

V=mgh

Where m is the mass of the system, g the gravitational acceleration, and h the height.

Another well-known and important potential energy is that for a spring that obeys Hooke's law.

V(x)=12kx2

The potential energy can otherwise be calculated from its definition; the value of potential energy at a point in space is equal to the negative of the Work needed to bring the object to that point from an arbitrarily chosen reference point in space, which is generally chosen so that the potential energy will never be negative.

This is a very broad question. The most broad answer i can give is that energy (mechanical, potential, chemical, etc...) can all be measured in Joulse. This is a unit for energy. Work and energy are the same, and work is defined as force (mass * acceleration) times a distance. The base unit of force is Newtons (1kg * (9.81 meters / second^2)) and for distance it is Meters, furthermore one joule of energy is equal to one Newton-Meter. There, now we know that one joule of energy is equivalent to moving 1 kilogram of mass at 9.81 m/s^2 over a distance of 1 meter; or any such manipulation of the

It seems, that you lack knowledge of Physics and Mathematics.

Go to Ramamurti Shankar “Fundamental Physics I” on You Tube and try to follow Prof. Shankar. I know no better teacher. Just watch the videos until you have understood all the remarks and expanations. Do not learn formulas by heart; do learn the logic of how to derive those formulas. Try to understand the Principles of Physics. Do not fool yourself. Work seriously and honestly. Use your logic and try to fix your logic. (I hope that you understand what I am talking about.)

With lesson 12–15 of “Fundamental Physics I” you will get the best introduction to Special Relativity Theory. The Lorentz Transformation are just beautiful and … everything follows from them. It will take you only 5 hours if you are smart.

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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

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