General Concept Question-work done by gravitational force.

In summary: 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!1. If you push a block along a HORIZONTAL plane, what is the work done on it by the GRAVITATIONAL force?
  • #1
integra2k20
35
0
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!
 
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  • #2
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?
 
  • #3
arildno said:
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?
 
Last edited:
  • #4
"since the horizontal component of the motion is zero?"

Eeh? The object MOVES horizontally.
In which direction does the gravitational force act?
 
  • #5
sorry i meant to say vertical component is 0 not horizontal
 
  • #6
integra2k20 said:
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 [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
 
  • #7
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?
 
  • #8
arildno said:
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!
 
  • #9
nrqed said:
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


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
 
  • #10
integra2k20 said:
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 [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!
 

Related to General Concept Question-work done by gravitational force.

1. What is work done by gravitational force?

Work done by gravitational force is the product of the force of gravity acting on an object and the displacement of the object in the direction of the force. In other words, it is the energy transferred to an object by the force of gravity as it moves through a distance.

2. How is work done by gravitational force calculated?

Work done by gravitational force can be calculated using the equation W = F * d * cos(theta), where W is work, F is the force of gravity, d is the displacement, and theta is the angle between the force and displacement vectors.

3. What are some examples of work done by gravitational force?

Examples of work done by gravitational force include objects falling from a height, water flowing downstream, and objects being lifted against gravity.

4. How does work done by gravitational force relate to potential energy?

Work done by gravitational force is closely related to potential energy. As an object moves against the force of gravity, it gains potential energy. This potential energy can then be converted into kinetic energy as the object falls back towards the ground.

5. Is work done by gravitational force always positive?

No, work done by gravitational force can be both positive and negative. If the force and displacement are in the same direction, work is positive. If they are in opposite directions, work is negative. This is because the direction of the force and displacement vectors affect the angle between them, which is a factor in the calculation of work.

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