B Work done when moving an object

[spaceman0x2a]

TL;DR Summary

How is initial velocity affected by a required displacement with a constant force, in the context of work?

I think I have an idea of this but I just want to be sure, and I also have some extra questions that I want to clear up.

The work done on an object is the dot product of the force vector with the displacement vector, Fd*cos(theta). Am I correct in that this is without regard to velocity? i.e. the trajectory of the object will be different for each configuration of force/displacement, that being the trajectory with the right starting velocity that just brings it from point A to B? (which I assume will be parabolic, given a constant force)

Next, how does this work with the possible starting velocities? I'm assuming that there may be multiple possible starting velocity angles, as long as you correct their magnitude - but not all 360 degrees. For example, the velocity could not have the same angle as the force vector, as no matter its magnitude it will always fly off into infinity. Nearby angles would not work either, as the velocity would converge towards the force vector without ever being able to reach the desired displacement. What is the relationship between this "no-go" zone and the angle between the displacement and force vectors?

 

[anuttarasammyak]

Work done on moving object during short time $\delta t$ is
$$\mathbf{F}\cdot\mathbf{v}\delta t$$
During time $[t_1, t_2]$ Integrating by time
$$\int_{t_1}^{t_2}\mathbf{F}\cdot\mathbf{v}dt $$
Does it have something to do with your question ?

A simple special case for explanation :
Say a object falls from height h to 0. Work done by gravity is mgh with no regard of fast or slow falling speed it has, e.g. free fall with any initial velocity, carried in bag with elevator, stairs, etc.

What kind of cases are in your thought ?

 

[PeroK]

TL;DR Summary: How is initial velocity affected by a required displacement with a constant force, in the context of work?

If a force has a constant magnitude and direction, then a particle has constant acceleration (magnitide and direction). The motion, as we know from projectile motion, is either constant acceleration in a straight line, or motion in a parabola.

I'm not sure what issue there is calculating the work done in each case?

The work done on an object is the dot product of the force vector with the displacement vector, Fd*cos(theta). Am I correct in that this is without regard to velocity?

In general $\theta$ is a function of time. And you have to integrate, as mentioned in the post above.

i.e. the trajectory of the object will be different for each configuration of force/displacement, that being the trajectory with the right starting velocity that just brings it from point A to B? (which I assume will be parabolic, given a constant force)

Next, how does this work with the possible starting velocities? I'm assuming that there may be multiple possible starting velocity angles, as long as you correct their magnitude - but not all 360 degrees. For example, the velocity could not have the same angle as the force vector, as no matter its magnitude it will always fly off into infinity. Nearby angles would not work either, as the velocity would converge towards the force vector without ever being able to reach the desired displacement. What is the relationship between this "no-go" zone and the angle between the displacement and force vectors?

I don't understand what you mean by this.

 

[Dale]

The work done on an object is the dot product of the force vector with the displacement vector, Fd*cos(theta). Am I correct in that this is without regard to velocity?

The difference between displacement and velocity in this context is the difference between work and power. For work $$dW=\vec F \cdot d\vec s$$ and for power $$P=\vec F \cdot \vec v$$ I find power to be more convenient because of the differentials involved for work. But they are closely related since $\frac{dW}{dt}=P$ and $\frac{d\vec s}{dt}=\vec v$.

For example, the velocity could not have the same angle as the force vector, as no matter its magnitude it will always fly off into infinity.

You can certainly have the velocity parallel to the force. The above formulas handle any orientation of force, displacement, and velocity.

 

[jbriggs444]

Next, how does this work with the possible starting velocities? I'm assuming that there may be multiple possible starting velocity angles, as long as you correct their magnitude - but not all 360 degrees. For example, the velocity could not have the same angle as the force vector, as no matter its magnitude it will always fly off into infinity. Nearby angles would not work either, as the velocity would converge towards the force vector without ever being able to reach the desired displacement. What is the relationship between this "no-go" zone and the angle between the displacement and force vectors?

It is hard to understand exactly what you are talking about here.

It sounds like you have a particular destination point in mind. A fixed target. You also seem to have a fixed force in mind (gravity, perhaps). But you will allow the starting velocity to vary, both in magnitude and direction.

Does that match what you had in mind?

Given that understanding, I do agree that there are "no go" starting velocity directions. If you aim too low, you will miss your target low, no matter how fast the bullet. If you aim directly into the ground, the only way you will hit the target is if the target is directly below you.

I also agree that there are multiple launch angles that will still allow you to hit the target. You can aim high with a slow bullet and allow it to arch downward to the target. Or you can aim low with a fast bullet and allow it to shoot toward the target on nearly a straight line.

Regardless, the work done by gravity on the bullet from launch to target will be the same regardless of the launch speed and angle. The velocity change will be larger for a slower bullet. But the energy change will be identical.

 

[A.T.]

Next, how does this work with the possible starting velocities? I'm assuming that there may be multiple possible starting velocity angles

For a given velocity magnitude, there can be up to two angles to hit a target. This was used in naval warfare to land 2 full salvos almost simultaneously on the target, by firing the first in the stepper trajectory (longer flight time).

But if you can vary angle and magnitude, then you have infinitely many solutions. To find the boundaries of that solution space, you need to derive the general solution und check for which parameter combinations the arguments to roots become negative etc.

 

[spaceman0x2a]

I also agree that there are multiple launch angles that will still allow you to hit the target. You can aim high with a slow bullet and allow it to arch downward to the target. Or you can aim low with a fast bullet and allow it to shoot toward the target on nearly a straight line.

Regardless, the work done by gravity on the bullet from launch to target will be the same regardless of the launch speed and angle. The velocity change will be larger for a slower bullet. But the energy change will be identical.

But if you can vary angle and magnitude, then you have infinitely many solutions. To find the boundaries of that solution space, you need to derive the general solution und check for which parameter combinations the arguments to roots become negative etc.

Yes, this is what I wanted to confirm. Thankyou everyone for bearing with me on this, I didn't explain what I was thinking of very well.

 

[Dale]

Yes, this is what I wanted to confirm. Thankyou everyone for bearing with me on this, I didn't explain what I was thinking of very well.

I am glad that they understood what you meant and were able to help.

I would like to be very clear that what they described was very specific to projectile motion and is not something relevant to a general definition of work. Please do not try to generalize their specific projectile motion comments to other scenarios.