# Homework Help: Acceleration, Force, and so on

1. Dec 1, 2003

### Captain_Insano

Ok, this seems fairly straight forward, but I can't get my head around it. You have an object (0.3kg) with a vertical (up; positive) velocity of 10 m/s. This object's velocity is supported by a rising piston. Once the piston stops it's movement, how do you determine whether the object has enough momentum to overcome gravity, and jump? What would be the "Force" required to stop this jump? OK, thanks in advance.

2. Dec 1, 2003

### Staff: Mentor

I don't know what you mean by "overcome gravity". When the piston stops moving, the object still has an upward velocity. It will continue upwards under the influence of gravity, like any object tossed in the air.

3. Dec 1, 2003

### Captain_Insano

I suppose this may be a better way to explain it. Lets say you have a skateboarder who goes up a ramp:
Skateboarder weighs 70kg
Total vertical height of ramp is 3 meters

One ramp has a linear ramp and a flat top, as follows:

........o o o o o o
......o
....o
..o

(Disregard the "."s, I had to place them in there as a place holder to get this to post correctly)

Disregarding all other forces (friction, air resistance, etc.), and asuming that the velocity up the ramp is constant, how much momentem is required to "jump" at the top of the ramp? As in, if the skateboarder were going slow, he would reach the top, and feel a little less in weight, but would not leave the surface of the flat top. However, if the skateboarder were to be going fast, he would become air borne.

What I'm trying to figure out is what velocity/momentum/whatever is required to break gravity and become airborne. (when will your feet leave the ground in an elevator)

Hope that clarifies, thanks for the response.

4. Dec 2, 2003

### Staff: Mentor

Think of it this way. The piston cannot stop instantly, it must be deccelerated. If its decceleration is greater than g (the acceleration due to gravity) then the object will separate from the piston. (Just like a tossed object separates from your hand.) To keep the object from leaving the surface, enough force must be applied to it to deccelerate it as much as the piston is deccelerating. (Hope this helps a little.)

5. Dec 2, 2003

### Captain_Insano

I understand that. However, in the case of the ramp, the lack of support of the object(skateboard) is instantaneous, right? You have a constant velocity of the skateboard, and thus, a constant vertical velocity. Then, at the top of the ramp, what happens? Do you automatically become airborne? And what proof, or formula, shows this? Thanks again.

6. Dec 2, 2003

### Staff: Mentor

First off, the skateboard slows down as it climbs the ramp. (You won't have a constant speed.) When it gets to the top, it will have (we presume) some non-zero vertical component of velocity left. Then, just like any projectile, it will take off. The center of mass of the skater plus board will follow a parabolic path just like any other projectile. Of course, if the velocity at the top is small, the amount of air time will be unimpressive.

Also, making things more interesting is the fact that the "skater + board" system is not a rigid body. The skater can pull up his legs, etc.

7. Dec 2, 2003

### Captain_Insano

OK, I think that answers my question. I understand what you mean whe you talk about not being a rigid body, but assuming that it is, it should become airborne, even if at some miniscule degree. So, what would be a good way to show "force" coming off the ramp? Momentum? Velocity? Thanks for all your help, as it has clarified the issue for me.