Block sliding down a ramp in a truck

[davidmigl]

Homework Statement

In this problem, I am supposed to find something wrong with a formula dealing with the problem. There is a block with mass m on a ramp of height h inside a box truck moving with constant velocity u. The angle of the ramp doesn't matter, as this problem will be using energy methods. v is the velocity that the block reaches on the floor due to the potential energy only being converted into kinetic energy. No friction is involved, and no relativistic effects come into play.

The block is released from rest, slides down the ramp, and then slides across the floor of the truck.

Homework Equations

From an observer inside the truck, the relevant equation for velocity once on the floor is $$mgh = \frac{mv^{2}}{2}$$, by conservation of energy. But observed from outside the truck, the block has initial velocity $$u$$, since it is moving with the speed of the truck. Let $$v_{f}$$be the unknown final speed. Then $$\frac{mu^{2}}{2} + mgh = mv_{f}^{2}$$. Once the block slides down the ramp, all the potential energy is converted into kinetic energy, so $$\frac{mu^{2}}{2} + \frac{mv^{2}}{2} = \frac{mv_{f}^{2}}{2}$$. Multiplying both sides by 2 and dividing by m leads to the conclusion that $$v_{f}^{2} = v^{2} + u^{2}$$, or $$v_{f}=\sqrt{v^{2} + u^{2}}$$.

Alternatively, it is given that $$mgh = \frac{mv^{2}}{2}$$. Add $$\frac{mu^{2}}{2}$$ to both sides to give $$mgh + \frac{mu^{2}}{2} = \frac{mv^{2}}{2} + \frac{mu^{2}}{2}$$. But $$mgh + \frac{mu^{2}}{2}$$ = \frac{mv_{f}^{2}}{2}[/tex], by conservation of energy. Substituting this back into the previous equation gives $$\frac{mu^{2}}{2} + \frac{mv^{2}}{2} = \frac{mv_{f}^{2}}{2}$$ and again it follows that $$v_{f}=\sqrt{v^{2} + u^{2}}$$.

The problem is that, intuitively, we know that $$v_{f}$$ should equal $$u + v$$. So somewhere in the previous discussion an error has crept in. What is that error?

The Attempt at a Solution

The conclusion possibly rests on the presumption that you can divide an object's speed up into parts and find the total kinetic energy by calculating the kinetic energy for each of the speeds, so long as all the speeds add up to the total speed. This is clearly false, as two bodies with equal mass traveling at speeds x and y will not have the same kinetic energy as an identical body traveling at speed x + y.

But the second alternative derivation arrives at the faulty conclusion via mathematical means and doesn't seem vulnerable to such an experimental/physical attack as above. Each of the premises seems sound, so it seems impossible to avoid arriving at the conclusion.

Something seems fishy about the initial velocity being in a different direction than the incline (i.e. the direction the block travels at first), but I don't know if I am on the right track here.

 

[Nate810]

The error in the previous discussion is that it assumed that the initial velocity u and the velocity v were in the same direction. This assumption is incorrect, as the block starts off with velocity u in the direction of the truck while the velocity v is in the direction of the incline. This means that the final velocity cannot be found by simply adding up the magnitudes of u and v. Instead, the relevant equation should be v_{f}^{2} = u^{2} + 2uv + v^{2}.

 

[baba89]

The initial velocity being in a different direction than the incline doesn't necessarily mean that the block will have a different final velocity than if the initial velocity was in the same direction as the incline.

One possible error in this problem is the assumption that the block will slide down the ramp without any friction. In reality, there will always be some friction present, which will affect the block's speed and final velocity. This could explain the discrepancy between the expected final velocity of u + v and the calculated final velocity of \sqrt{v^{2} + u^{2}}. Additionally, the assumption that the potential energy will be converted entirely into kinetic energy may not hold true in the presence of friction.

Another possible error is the use of the equation mgh = \frac{mv^{2}}{2}. This equation assumes that the block's potential energy at the top of the ramp is the same as its kinetic energy at the bottom, which may not be the case if the block is moving with the truck's velocity. The equation should instead take into account the block's initial kinetic energy due to the truck's velocity.

Furthermore, the use of the equation \frac{mu^{2}}{2} + \frac{mv^{2}}{2} = \frac{mv_{f}^{2}}{2} assumes that the block's initial kinetic energy due to the truck's velocity is equal to its final kinetic energy on the floor of the truck. This may not be accurate as the block may lose some of its initial kinetic energy due to friction or other factors.

In conclusion, the main error in this problem may be the assumption of no friction and the use of simplified equations that do not take into account the block's initial kinetic energy due to the truck's velocity and potential energy on the ramp. This results in a discrepancy between the expected final velocity and the calculated final velocity, highlighting the importance of considering all factors in a real-world scenario.