Work + Energy OMG!! Please HELP ME.?

Question

Hi. I'm not a beginner in Physics and I have a good understanding of work/ energy.
but this confuses me:


When I push an object directly upwards, the work I do is obviously mgh. BUt when I push an object up and at angle not = 90 degrees, (I KNOW the work done is ALSO mgh) but it confuses me how I get there. )
Wouldn't the work done have to be equal to the force (in the direction I'm pushing) * distance this force (which is at an angle) acts?
Why does mgh work when h just accounts for the vertical displacement and not the horizontal component.
I understand that an energy transformation is taking place, and Ep is increasing obviously...but if I push at a constant speed at an angle upwards, that means mg is balanced with the Normal force and Applied force y components right? but then isn't there a horizontal component that is a force and a horizontal displacement obviously , so what about this work??

Can someone explain a definition of work that would work for all cases ?

Answer 1

In physics, mechanical work is the amount of energy transferred by a force. Like energy, it is a scalar quantity, with SI units of joules. Heat conduction is not considered to be a form of work, since there is no macroscopically measurable force, only microscopic forces occurring in atomic collisions. In the 1830s, the French mathematician Gaspard-Gustave Coriolis coined the term work for the product of force and distance.

Positive and negative signs of work indicate whether the object exerting the force is transferring energy to some other object, or receiving it. A baseball pitcher, for example, does positive work on the ball, but the catcher does negative work on it. Work can be zero even when there is a force. The centripetal force in uniform circular motion, for example, does zero work because the kinetic energy of the moving object doesn't change. Likewise, when a book sits on a table, the table does no work on the book, because no energy is transferred into or out of the book.

When the force is constant and along the same line as the motion, the work can be calculated by multiplying the force by the distance, W = Fd (letting both F and d have positive or negative signs, according to the coordinate system chosen). When the force does not lie along the same line as the motion, this can be generalized to the scalar product of force and displacement vectors.

Answer 2

A force -in the direction of motion - performs work.

Work is done against a force, when force is -opposite to the direction of motion.

Work is always measured by the product of force and the distance along which the force acts.

In the case of lifting objects from the ground to a certain height h vertically the work needed is force x displacement = mg (force) x h (displacement).


The next important point in the case of gravitational pull is ‘whatever is the path by which you move an object from the ground to the height h, the total work is always mgh’.
The path may be any complicated or zigzag way.

In the case of an inclined plane if you move the object along the plane the work done is the force along the plane x the distance along the plane.

I f L is the length and F is the force then work done is F x L.

[Note that I have not used the vertical distance]

From the geometry of the inclined plane, L = (H/sinθ).

Therefore the work done can be written as F x (H / sinθ).

This can be written as (F/ sinθ) x H = F”x H

Thus the work done is the same if a force F” = (F/ sin θ) acts on the body and lifts it through the height H.

Thus to calculate the work done we can either use the vertical height H or the slanting length L. But the forces are different.

If F” is the force we use to calculate the work through distance H,
we have to use the force F = F” sin θ for calculating the work through the length L.

-------------------------------------------------------------------------------------------------

About the components of force, it is wrong to say that normal force balances the weight in an inclined plane.

Mg acts vertically down ward. Resolve this force into two components one along the inclined plane (mg sin θ) pulling the object down along the plane and the other one
(mg cos θ) pushing the object into the inclined plane. The normal reaction is (mg cos θ pulling the object away from the inclined plane at angle θ to the vertical.

The only unbalanced force is mg sin θ pulling the object down the inclined plane,

In order to move the object up the inclined plane we apply a force of mg sin θ and move it through a distance L.
.
Workdone by this force is mg sin θ x L = mg x L sin θ.

But L sin θ = H and the work done = mg H or mg sin θ x L

Answer 3

The work done in pushing the object up a ramp is still mgh, and it is still F·d (where F is the force in the direction you're pushing, and d is the length of the ramp).

The short explanation is: when you go up the ramp instead lifting directly vertically, the increase in the distance (d vs. h) is exactly offset by a corresponding decrease in the force (F vs. mg).

Here's the math:

Say the ramp forms angle θ with the horizontal. You can see by drawing a diagram that there's a relationship between h (the height of the high end of the ramp) and d (the length of the ramp). Namely:

h/d = sinθ

Remember that for later. Now, consider the forces acting on the object (ignore friction for now). They are:

• Object's weight (mg). Call it Fg.
• Normal force (due to ramp). Call it Fn.
• The push applied by you. Call it Fa.

Since we're postulating that the object is moving at a constant speed, it must be that the above 3 forces are in balance; i.e. their vector sum is zero.

If you draw a diagram, you'll see that this implies that the magnitude of Fn is mg•cosθ; and the magnitude of Fa is mg•sinθ.

For the purpose of calculating work, the only force that counts is the one that moves in the direction of motion: namely Fa. So the work done in pushing the object up the ramp is:

Fa•d
= (mg•sinθ)•d

But remember that h/d = sinθ. That means the work is:

(mg•sinθ)•d
= mgh

Answer 4

The answer is yes, there is a unified equation that works for all cases and it is the one preferred by physicists but it gets very complicated and uses the dot product and vectors.

It is literally, the "force vector dot the direction vector".

= F•D

wikipedia actually describes it well:
http://www.wikipedia.org/wiki/Mechanical_work

There is also a more complicated one that is based on calculus:

W = ∫F•ds

So this is the preferred one by physicists but bit is definately overkill for simple mgh problems. good luck.

Answer 5

hi check this link its good




http://workathomeoutsourcingjobs.blogspot.com/




.