Seemingly easy physics question

[Zorric]

Homework Statement

A .5kg apple is dropped off the top of a 15m tall building. Assuming there is no air resistance, what will the apple's speed immediately before it strikes ground be?

Homework Equations

U=m g h

The Attempt at a Solution

(.5kg)(9.8m/s^2)(15m) = 73.5 m/s?

Just doesn't seem that simple, though.

 

[Nathanael]

(.5kg)(9.8m/s^2)(15m) = 73.5 m/s?

On the left side, you have units of energy (Joules) but on the right side, you have units of speed, so it can't be equal.

EDIT:

U=m g h

What does the "U" mean in this equation?

 

[Zorric]

On the left side, you have units of energy (Joules) but on the right side, you have units of speed, so it can't be equal.

EDIT:

What does the "U" mean in this equation?

gravitational potential near the surface of Earth according to https://www.physicsforums.com/showpost.php?p=3820479&postcount=7

This seems like the most logical equation to use granted the only known data is the height, gravitational force, and mass.

 

[Nathanael]

the only known data is the height, gravitational force, and weight.

I'm sure you meant to say mass :)

This seems like the most logical equation to use granted the only known data is the height, gravitational force, and weight.

Would the mass of the object even matter? If you drop two objects of different mass from the same height, will they hit the ground at different times/speeds? (Ignoring air resistance, like in the problem.)

This seems like the most logical equation to use granted the only known data is the height, gravitational force, and weight.

There is a bit more information, it's just slightly hidden:
"A .5kg apple is dropped off the top of a 15m tall building."
If it is dropped, then what do you know about it's initial speed?

 

[Zorric]

I'm sure you meant to say mass :)


Would the mass of the object even matter? If you drop two objects of different mass from the same height, will they hit the ground at different times/speeds? (Ignoring air resistance, like in the problem.)


There is a bit more information, it's just slightly hidden:
"A .5kg apple is dropped off the top of a 15m tall building."
If it is dropped, then what do you know about it's initial speed?

Indeed I did mean mass. :) You are right about the two objects dropping at the same time, if it is dropped than the initial speed will be 0 m/s.

 

[Nathanael]

Indeed I did mean mass. :) You are right about the two objects dropping at the same time, if it is dropped than the initial speed will be 0 m/s.

So you know the initial speed, and you know the way in which the speed is changing (the acceleration), so are you able to draw a graph of what the speed would be at any time?

In other words do you know the speed as a function of time? (assuming when it is dropped t=0)

 

[Zorric]

So you know the initial speed, and you know the way in which the speed is changing (the acceleration), so are you able to draw a graph of what the speed would be at any time?

In other words do you know the speed as a function of time? (assuming when it is dropped t=0)

Not exactly sure, I am sure you can derive something using the initial velocity, just not sure.

 

[Nathanael]

Not exactly sure, I am sure you can derive something using the initial velocity, just not sure.

The acceleration is 9.8 m/s per second. Meaning every second the speed increases by 9.8 m/s

Think of an XY coordinate system, except the X axis will be the Time axis and the Y axis will be the Speed axis

At time zero, speed is zero. At time 1 second, the speed is 9.8 m/s

Let's generalize it. What would be the speed after "T" seconds?

 

[Panphobia]

I think you guys are over thinking it. You have obviously learned about work and energy since you know mgh. So then you should know that the total energy before the event has to equal the total energy after the event, and total energy = K + U. So this would be K + U = K' + U' where K'+U' is the energy after the event. You already know that potential energy U = mgh and that kinetic energy K = (1/2)*m*v^2, so just plug those in and see what you get.

The equation with everything subbed in is.
mgh + (1/2)*m*v^2 = mgh' + (1/2)*m*v'^2

where h' is the height after the event, and v' is the speed after the event.
So you have this equation, and all you need to do is plug in the numbers.

(.5)*(9.8)*(15)+(1/2)*(.5)*(0) = (0.5)*(9.8)*(0) + (1/2)*(.5)*v'^2

(2)*(9.8)*(15) = v'^2

v' = sqrt(2*9.8*15)

 

[Nathanael]

I think you guys are over thinking it. You have obviously learned about work and energy since you know mgh. So then you should know that the total energy before the event has to equal the total energy after the event, and total energy = K + U. So this would be K + U = K' + U' where K'+U' is the energy after the event. You already know that potential energy U = mgh and that kinetic energy K = (1/2)*m*v^2, so just plug those in and see what you get.

I know that this is the simplest solution, however I assumed he had not yet learned about Kinetic Energy and Conservation of Energy (because if he did know about it, he would have likely realized this solution when I told him that the units on the left are that of energy)

I wasn't overthinking it, I was walking him through the logic of an alternative solution that only involves an understanding of the relationship between distance/velocity/acceleration. (Because you can solve this sort of problem before you've learned about conservation of energy...)

There's no point in just giving him the answer if he didn't partially figure it out himself. No learning happens from being given the answer.
P.S. I think you should only use the Conservation-of-Energy solution if you already thoroughly understand the alternative solution, because the alternative solution is "more true" or "more fundamental" in the sense that it involves the least amount of prerequisite information (you don't need to know any "Laws" to find the answer.)

The Conservation-of-Energy solution is merely a convenient short-cut. It doesn't create understanding.

 

[Panphobia]

Yes you can solve it without conservation of energy, but he was partially using conservation of energy, so I thought that is what he needed.

 

[Nathanael]

Yes you can solve it without conservation of energy, but he was partially using conservation of energy, so I thought that is what he needed.

But he said his reason for using the GPE equation was simply that it involved all the variables he was given.

That means he wasn't actually thinking about conservation of energy.

 

[Zorric]

thanks for the help!

Thanks for the help guys, not sure what the confusion was between you two, I have learned about kinetic energy and whatnot, just trying to brush up on these problems. Also, that was the equation I was trying to remember @panphobia and when you mentioned it, I automatically knew that was how I did it previously

 

[Nathanael]

Thanks for the help guys, not sure what the confusion was between you two, I have learned about kinetic energy and whatnot, just trying to brush up on these problems.

Have you learned about kinematics equations? It is from the understanding of kinematics equations (of constant acceleration) that you would solve this problem.

If you can't solve it that way you should brush up on the relationship between acceleration/velocity/displacement (specifically when acceleration is constant).

If you can solve it with kinematics, then you could use conservation of energy to solve it more easily by realizing that all of the GPE will be converted into KE and therefore: $mgh=\frac{mv^{2}}{2}$
(You'll notice the mass cancels out on both sides, which reflects what I said about that piece of information being unneccessary and irrelevant.)