# Increased angle should lead to decrease height

I'm having trouble understanding the answer for the following problem:

• The diagram below represents a 10-kilogram object at rest at point A. The object accelerates uniformly from point A to point B in 4 seconds, attaining a maximum speed of 10 meters per second at point B. The object then moves up the incline. The object comes to rest at a vertical height of S (point D) when $\theta$ = $30 ^\circ$. If $\theta$ were increased to $40 ^\circ$, the object would come to rest at a vertical height at: Less than S, Greater than S, or the Same?* Given the restraints in the problem, I expected the answer to be less than S, because all conditions, including the velocity, and acceleration of the box stays the same in both problems. However, an increased angle should intuitively lead to a decrease in the vertical position where the box comes to a rest. However, the answer key states otherwise: the height remains the same.

• "object come stores t"? Ah, "comes to rest".
– user137289
Feb 4, 2018 at 19:57
• Your intuition may be right. The object would not move up a vertical wall. But this is a textbook physics problem, not the real world. The only thing that counts here is potential energy.
– user137289
Feb 4, 2018 at 20:04

Yes, @Pieter is right, the textbook physics is definitely not the real world. But, there is one thing that you have to take into account.

When $\theta$ passes $45^\circ$, the force of friction does, in fact, decrease because $\cos\theta$ decreases as $\theta$ increases past $45^\circ$. Thus, it is possible that the friction force decreases i.e. the block can move farther up the incline before friction takes away the energy.

• The problem does not mention friction. In this textbook problem, I would assume this should be solved disregarding friction. (Although they did not even draw wheels under that box!)
– user137289
Feb 4, 2018 at 20:31

If one ignores friction (say the surface is very smooth) then this is a classic problem on conservation of energy.

The total energy at the start of the motion is entirely kinetic: $$E=\frac{1}{2}mv^2 \tag{1}$$ whereas, when the box reaches it maximal height, the velocity will be $0$ (otherwise it would continue going up) and at that precise moment all the energy is potential: $$E=mgh \tag{2}$$ By conservation of energy, (1) and (2) are equal so $$\frac{1}{2}mv^2=mgh \qquad \Rightarrow \qquad h= \sqrt{\frac{v^2}{2g}}\, .$$ Besides not depending on how you reach the maximal height $h$, this maximal height is also independent of the mass of the box, i.e. a $5$kg box would reach the same height as a $10$kg box if one ignores friction.

Of course in a more realistic case there would be some friction but, in the case of a roller coaster for instance, the friction is sufficiently small so that the cart always reaches maximum height irrespective of the mass of the cart and the people in it, so neglecting friction is not de facto unrealistic, at least in some situations.