Effect of gravitational force

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mgh force

Will there be any change in the required force if one pulls an object horizantally (perpendicular to earth) instead of vertically (against gravitational force)?
To make my question clear, here is an eg.
Suppose I'm pulling a heavy object against gravity. Lets say I spend energy 'x'. now if I pull the same object horizantally (somehow, assume that i've made a setup like that), then say energy spent is 'y'. Will there be any difference between 'x' and 'y'? If so, why?

Please explain.
Thanks,
 

effect of gravitational force on displacement

They are totally different. If you pull an object vertically, work is done to against the gravity force, i.e. E = mgh. However, if you pull the same object horizontally, work is done to against the friction between the object and the horizontal surface e.g. table, or ground floor, E = Friction force x distance . Thus, as you mentioned, the energy you spend on pulling the object "x" and 'y" are not equal.
Hoping the above explanation is helpful to you.
 

fxdxcos

elow..
Yes there is a big difference between that..
one difference is the work done by pulling that object..
as we all know work done is equal to
W=Fxd=FdsinΦ,
wherein Φ is the angle between the F and d..
as you pull the object horizontally Work is done..
but if yoy pull the object vertically, the Force you apply is parallel to the covered distance means the work done is Zero...


-reu
 

force mgh

I agree with the comments above. I would offer only one more comment to make it simpler.

Work = Force times distance.

Think about the force you are pushing against in each case.
Pushing vertically => pushing against gravitation energy => the energy result of your transaction is inputting work and increasing the gravitational potential energy of the object.

Pushing horizontally => pushing against friction => the energy result of your transaction is inputting work and creating heat through friction.

Also if the object is not at the same velocity before and after pushing, a certain component of the force times distance input energy is associated with accelerating the mass. This work is converted to kinetic energy.
 

w=fdsin

but if yoy pull the object vertically, the Force you apply is parallel to the covered distance means the work done is Zero...

This is not correct.
 

pull object horizontally

but if yoy pull the object vertically, the Force you apply is parallel to the covered distance means the work done is Zero...

This is not correct.
Click to expand...

If I remember correctly during my school time, work done and energy is different, though they share the same units.

Work Done relates to displacement. No displacement, no work done.
Specifically to the example, if you pull the object vertically, work done=0. However energy on object is not zero.
 

if you pull the object vertically, you are doing work against gravity since you are displacing it:

W=-mgh

According to your definition, the angle between the force and displacement is 180° and the cos of this angle is -1, not zero.

Rather, in the hor. direction, the work done against gravity is zero, but there is some work done against some other forces like frictional force.

Added after 7 minutes:

W=Fdsin () formula is not correct, it is W=Fdcos(), in vector notation:

W=F•d

And, work done is not entirely different than energy, rather energy has something to do with work, it is the ability to generate work against some force. When you do work against gravity the object has grav. potential energy=mgh, which can set it to motion in the direction of gravity.
 

Work done vertically is different.This is done against gravcitational force which is conservative(work done does not depend on path).Whereas the work done against friction(non conservative) is dependent on path.
 

that's ok
 

W=-mgh,



that's not an equation of work..that's the equation for Potential Energy aquired when your applying force against the gravity..


-reu
 

rkenuox_ said:
W=-mgh,



that's not an equation of work..that's the equation for Potential Energy aquired when your applying force against the gravity..


-reu
Click to expand...

I have given a neg. sign to indicate it is the work done when displacement is opposite to the direction of the force. When you do work against a conservative force field, that work is stored as potential energy. Energy and work has the same units. Because you are having a dispalcement against gravity here, work is obviously done. Otherwise from where you get the energy? If still in doubt go for some school physics books.
 

Here's an alternative explanation: You look at the initial and final states of the system. Although this method doesn't really work all the time, but its good here.

See at first the initial states are the same. But the final states are different - if you lift it up it now has gpe, if you pull it sideways it only generates that little bit of heat due to friction.
 

well friend,u no this is just a delusion.gravity is same.
yet quantitatively the two r different.vertical motion makes u spend Mgh amount of energy while as horizontal motion makes u spend enerry depending on various factors as friction,the manner in which force is applied.but the point to remember is that gravity works in vertical only n any displacement in vertical direction dissipates energy (hence work) against gravity,rest may be used in overcoming friction etc.
 

There will be no energy difference, because when you pull it also horizotally you make no work and thus spend no energy.

Work=Fxdxcos(theta)

Theta= Angle between force and displacement vector

In horizontal theta is 90°.

So no work.
 

i will say one thing, but i am not sure whether it is right.

when u move a body against gravity you can intuitively make sense that a work is being done as h increases, at every point you need to put extra effort to take it higher.

f=mgh(potential energy).

incase, you move it horizontally, still you would move it at a constant height h only. so surely you would need energy to maintain that object at that height.

AMRITH.S.


 

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