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).

 

[PJC01]

I can provide a more accurate response to your questions about the acceleration due to gravity and the movement of a cart on an inclined plane.

First, let's define some terms. The acceleration due to gravity is a constant value, denoted by "g," and is equal to 9.81 m/s^2. This value represents the rate at which objects accelerate towards the Earth due to the force of gravity.

Now, let's address your first question: What would be the acceleration of the cart right after it is launched?

As you correctly stated, the acceleration of the cart would be 9.81*cos(theta) m/s^2. This is because the cart is being launched at an angle theta, and only the component of gravity acting in the direction of motion (the cosine component) will contribute to the acceleration.

Your second question is: What would be the acceleration of the cart at its highest point?

At the highest point of the cart's trajectory, the acceleration would be 0 m/s^2. This is because at this point, the cart has reached its maximum height and is momentarily at rest before it begins to fall back down the inclined plane. The force of gravity is still acting on the cart, but it is balanced by the normal force from the inclined plane, resulting in a net force of 0 and therefore, 0 acceleration.

Finally, your third question is: What would be the speed of the cart when it returns to the point of launch?

To answer this question, we need to consider the conservation of energy. At the point of launch, the cart has a certain amount of potential energy due to its height on the inclined plane. As it moves back down the plane, this potential energy is converted into kinetic energy, resulting in an increase in speed. The speed of the cart at the point of launch would be equal to the speed it had at the bottom of the inclined plane, which would depend on the angle of the incline and the distance traveled.

I hope this helps to clarify the concept of acceleration due to gravity and the movement of a cart on an inclined plane. Remember, the key is to consider the forces acting on the cart at different points in its trajectory and how they contribute to its acceleration and speed.