Acceleration due to gravity - cart pushed up an inclined air track (discussion)

[shawli]

Say I launch an object (a cart) up a frictionless inclined plane at an angle of 'theta' to the horizontal.

Question: What would be the acceleration of the cart right after it is launched?

My guess at the answer: 9.81*cos(theta)
since the cart is launched at an angle.
BUT I'm not sure and I also think that acceleration might be constant at -9.81m/s² all throughout the cart's movement up and down the inclined plane because there is no force aside from gravity acting on the cart as soon as it is launched.

Question: What would be the acceleration of the cart at its highest point?

My guess at the answer: Not sure...

Question: What would be the speed of the cart when it returns to the point of launch?

My guess at the answer: Seems like the speed would be the highest here because v_f=v_i+a*t and since t is at its greatest magnitude once the cart has returned back to the point of launch, the final velocity should be greatest as well.


I'm really fuzzy on my physics concepts because I haven't seen this stuff in awhile -- I hope someone can help refresh/explain some of these ideas to me! Thank you for reading :)

 

[Ken G]

My guess at the answer: 9.81*cos(theta)
since the cart is launched at an angle.

Correct.

BUT I'm not sure and I also think that acceleration might be constant at -9.81m/s² all throughout the cart's movement up and down the inclined plane because there is no force aside from gravity acting on the cart as soon as it is launched.

That would be true if there really were no other forces on the cart. But is the cart touching anything that could put a force on it? When does the thing it is touching exert a force, and what constraint does that force make the cart satisfy?

Question: What would be the acceleration of the cart at its highest point?

Does the angle of the incline change? Why would the forces change?

Question: What would be the speed of the cart when it returns to the point of launch?

My guess at the answer: Seems like the speed would be the highest here because v_f=v_i+a*t and since t is at its greatest magnitude once the cart has returned back to the point of launch, the final velocity should be greatest as well.

Careful, the sign of a is negative here, so the magnitude of the speed actually goes down before it goes up. Your conclusion is correct, but you can prove that it's correct and even get the value of the speed, simply by asserting conservation of energy. Can any of the energies be different at the end of the problem as from the start, and if so, how could that be possible and still conserve energy? If all the energies are the same, what does that tell you about the final speed?

 

[shawli]

Correct.
That would be true if there really were no other forces on the cart. But is the cart touching anything that could put a force on it? When does the thing it is touching exert a force, and what constraint does that force make the cart satisfy?

Are you referring to the normal force and force of friction here? I'm not certain, but it seems like those would become irrelevant because they should be acting in the same magnitude whether the cart is going up or going down.

Correct.
Does the angle of the incline change? Why would the forces change?

Ah, so the acceleration stays constant because the angle of incline does not change? Thank you!

Careful, the sign of a is negative here, so the magnitude of the speed actually goes down before it goes up. Your conclusion is correct, but you can prove that it's correct and even get the value of the speed, simply by asserting conservation of energy. Can any of the energies be different at the end of the problem as from the start, and if so, how could that be possible and still conserve energy? If all the energies are the same, what does that tell you about the final speed?

Hmm, this should mean that the initial velocity should be the same as the final velocity eh? What I can't wrap my head around exactly is the implication of the cart being "launched". Doesn't this mean that the launch of the cart has some additional energy that return of the cart to its initial point would be missing?


Aside: This is a stupid question, but I just want to double check -- the position-time graph for this cart-on-an-incline will look like an upside down "V" right? The velocity-time graph should be a "U" and the acceleration-time graph should be a horizontal line?

 

[Ken G]

Are you referring to the normal force and force of friction here?

Just the normal force-- in these kinds of problems, friction is usually ignored, unless directly specified.

I'm not certain, but it seems like those would become irrelevant because they should be acting in the same magnitude whether the cart is going up or going down.

They will indeed, by that does not make them irrelevant. Still, there doesn't seem to be a problem-- your first answer was correct.

Ah, so the acceleration stays constant because the angle of incline does not change? Thank you!

Yes.

Hmm, this should mean that the initial velocity should be the same as the final velocity eh?

Quite so.

What I can't wrap my head around exactly is the implication of the cart being "launched". Doesn't this mean that the launch of the cart has some additional energy that return of the cart to its initial point would be missing?

Yes, but we generally imagine starting the problem after the launch has happened, so the "initial" speed is post-launch.

Aside: This is a stupid question, but I just want to double check -- the position-time graph for this cart-on-an-incline will look like an upside down "V" right? The velocity-time graph should be a "U" and the acceleration-time graph should be a horizontal line?

Not quite-- it is the magnitude of speed graph that is a right-side up V (constant acceleration downward along the incline). That means the position-time graph is the upside-down U (since V is the magnitude of the slope of U). The acceleration is flat, since that is the slope of V when V is the magnitude (so the actual velocity goes negative rather than coming back up like a V).