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 isn't true and i don't 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]

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?

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]

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).

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.

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 isn't true and i don't understand why or what it should be. Thanks!

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 W = F d cos \theta
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 \theta= 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 Work_{gravity} = - mg d.

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 mg y_i - mg y_f 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]

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?

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 isn't 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]

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 W = F d cos \theta
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 \theta= 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 Work_{gravity} = - mg d.

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 mg y_i - mg y_f where "i" and "f" stand for initial and final y positions. This form is probably easier to work with.

Hope this helps

Patrick

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

 

[nrqed]

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

Ok. Then think of mg y_i - mg y_f. 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 mg y_i - mg y_f is only applicable if the y-axis points straight up )

Best luck!