Car starts with some kinetic energy

[MIA6]

Homework Statement

This is a not hard question but you may imagine a image in your mind since I can't provide you the diagram: A 10-kilogram object at rest at point A. The object accelerates uniformly from point A to point B in 4 seconds. (A and B is a straight line. A...B) The object then moves up the incline. (Neglect friction. So at the point B on there is the incline,the end point of the incline is D. so the incline is BD. Imagine a right triangle, BD is the hypotenuse, the vertical height is S) Question: The object comes to rest at a vertical height of S (point D) when the angle of incline is 30, If it increases to 40, the object would come to rest at a vertical height 1) less than S , 2) greater than S, 3) equal to S

The Attempt at a Solution

I chose 2) since if you draw a 40 degree angle, and at the same time you need to keep BD the same distance, so the vertical distance is larger than before. But my teacher said it was 3). I don't get it. Hope you can give me some idea.

 

[G01]

The answer is indeed #3.

Think conservation of energy. The car starts with some kinetic energy at the bottom of the incline and loses it as it goes up the incline. BD, will not remain the same distance from each other. In the situation with the higher slope, B and D will be closer to each other, since the vertical height can't change, due to conservation of energy. What is the equation describing conservation of energy here? Does the angle of the slope play any part in it?

 

[MIA6]

The answer is indeed #3.

Think conservation of energy. The car starts with some kinetic energy at the bottom of the incline and loses it as it goes up the incline. BD, will not remain the same distance from each other. In the situation with the higher slope, B and D will be closer to each other, since the vertical height can't change, due to conservation of energy. What is the equation describing conservation of energy here? Does the angle of the slope play any part in it?

Why the vertical height can't change because of conservation of energy. In the second case which the angle is 40 is independent of the first case 30 degree angle. So their conservations of energy need not to be equivalent. 1/2mvi^2+mghi=1/2mvf^2+mghf. The problem for me now is h,(h=S). How can I find out?

 

[Doc Al]

I chose 2) since if you draw a 40 degree angle, and at the same time you need to keep BD the same distance, so the vertical distance is larger than before.

The distance from B to D will change depending on the angle.

Why the vertical height can't change because of conservation of energy. In the second case which the angle is 40 is independent of the first case 30 degree angle. So their conservations of energy need not to be equivalent. 1/2mvi^2+mghi=1/2mvf^2+mghf. The problem for me now is h,(h=S). How can I find out?

But vi, vf, and hi are the same for both inclines. So how can hf be different?

 

[MIA6]

I thought hi=hf, it doesn't change. Moreover, if the angle is 40, then the question asks you the new height I think. So how do you know hi is the same for both inclines.

 

[Doc Al]

The initial height (hi) is the height at the bottom of the incline (at point B), which is the same for both inclines. hf is the final height of the object after it moves up the incline (at point D): hf > hi in all cases.