General Concept Question-work done by gravitational force.

integra2k20

OK, I'm having trouble with this and can't seem to find an explanation for it. So To start off, i understand gravitational force, which is always just m*g (mass * -9.8 m/s).

Now, when an object is moved by a force either horizontally or up an incline, etc. how do you figure out the work done on the block by this gravitational force? I thought it would just be W = f*d but apparently this isnt true and i dont understand why or what it should be. Thanks!

arildno

1. If you push a block along a HORIZONTAL plane, what is the work done on it by the GRAVITATIONAL force?

2. Thus, what distance (component) is relevant in order to calculate the work of the gravitational force?

integra2k20

if its a horizontal plane would it be zero since the horizontal component of the motion is zero?

arildno

"since the horizontal component of the motion is zero?"

Eeh? The object MOVES horizontally.
In which direction does the gravitational force act?

integra2k20

sorry i meant to say vertical component is 0 not horizontal

nrqed

True as long as you stay relatively close to the surface of the Earth. Otherise you need to start using the universal law of gravitation...But that's irrelevant for you now.

The general definition of the work done by any force when an object is moved along a straight line and the force is constant (in both direction and magnitude) is [tex] W = F d cos \theta [/tex]
where "d" is the distance over which the object is moved and the angle is between the direction of the force and the direction in which the object has been moved.


If the object is moved horizontally and you want the work doen by gravity, then [itex] \theta [/itex]= 90 degrees so the work done by gravity is zero. If the object is moved along straight up, then the angle is 180 degrees (force of gravity is down, the motion is upward) and we get [itex] Work_{gravity} = - mg d[/itex].

You can show that if you use a y axis that is vertical and with the positive y direction pointing up, you always find that the work done by gravity is equal to [itex] mg y_i - mg y_f [/itex] where "i" and "f" stand for initial and final y positions. This form is probably easier to work with.

Hope this helps

Patrick

arildno

Correct!
So, ask yourself:
Can the horizontal component of the distance traveled ever be relevant in calculating the work done by the gravitational force?

Secondly, how does this relate to the work done by g.f, if the object moves on an incline?

integra2k20

ok...so, basically, if the motion is strictly horizontal, gravitational force does NOT do any work. this didnt make sense to me because i would still assume gravity was pulling down, but i guess since the object isnt moving as a result it is not techincally doing any work.

as for an incline, i would have to assume that if an object moves a certain distance up the incline, the distance over which gravitational force would be acting would be only the horizontal part. It all makes sense now! so if it moved 2 meters up a 30 degree incline for example, the distance that i would need to use to calculate work done by GRAVITY would be 2sin(30) or 1 meter. Thanks everyone for the help!

integra2k20

some of that was a little too in depth for me...this is my first year studying physics, but the Fdcos(theta) will definetely prove a big help in the future, im going to print that out and keep it in my notes

nrqed

Ok. Then think of [itex] mg y_i - mg y_f [/itex]. The part of the distance that enters in the work done by gravity is only the change of vertical position. As you said in your incline example, if an object moves along an incline 2.00 meters long at an angle of 30 degrees above the horizontal, only 2.00 sin(30) = 1.00 meter will matter in the work done by gravity. If the object moved *down*, the work done by gravity will be positive and it will be negative if the object moves up.

Only one thing to be careful about: the formula [itex] mg y_i - mg y_f [/itex] is only applicable if the y axis points straight up )

Best luck!