[RESOLVED] Quick Easy Energy/Physics Problems

[Zach_VB6]

Here's what I'm given (not understanding this much):

A 20kg mass is on the edge of a 100m high cliff.
a) If the mass falls from the cliff, what will kinetic energy be just before impact?
b) If the mass falls from the cliff, what will be the speed of the mass just before impact?

Thank you guys so much :)

(an explanation would help a lot too)

[chrisman10]

v2*100*9.8 = 44.0
That is V
The reason you need the Velocity is to find the K.E
So KE is
19600.0 because 1/2*2*44^2

[Zach_VB6]

velocity = square root of (2 * 10 * 100) = sqr root of 2000
KE = (mv^2)/2 = (20 * 2000) / 2 = 10 * 2000 = 20,000

Am i right?

[chrisman10]

EK = (1/2)mv2

EK= .5*20*44^2 = 19360

Correction its not 19600 its EK= 19360 .

Your first part is right but second part is wrong. You have V correct.

[Zach_VB6]

(The square root of 2000) ^ 2 = 2000...

[chrisman10]

Yeap your right but you have to square root is so its 44.0. :P

[chrisman10]

velocity = square root of (2 * 10 * 100) = sqr root of 2000
KE = (mv^2)/2 = (20 * 2000) / 2 = 10 * 2000 = 20,000

Am i right?

Sorry I did not see you were rounding ;p its prefect.

[jemidiah]

I have some free time at the start of the summer and figured I'd write something slightly more in-depth about this problem, since it interests me. If you find it interesting or illuminating, that's great, though parts of this post will be beyond the scope of an introductory mechanics course. For those looking at a wall of text and losing interest, at the end I give a very high-level overview of how the double pendulum problem is solved. A picture of one possible path is included below.

http://upload.wikimedia.org/wikipedia/en/4/45/Double-compound-pendulum.gif


There are two major ways of doing this problem. The first involves forces in intermediate steps, and the second involves energy. They are essentially the same approach, viewed from two different mathematical perspectives: they are equivalent through path independence of line integrals in the conservative gravitational vector field. (This path independence implies the existence of a potential function, here called gravitational potential energy, which can be used to evaluate these line integrals more easily. MathWorld has a very terse description (http://mathworld.wolfram.com/ScalarPotential.html) of a scalar potential, which is what the gravitational potential in this case is.)

First, try doing this problem using forces. Suppose the object were thrown down very quickly at the start; the final velocity would depend on the initial downward velocity, which we weren't given. So, let's assume the object starts falling off the cliff with zero initial velocity. That is, v(0) = 0, where v(t) = velocity of object at time t, and t=0 means the instant the mass starts falling. Drawing a free body diagram, you can very easily convince yourself that the object experiences only a downward force due to gravity of m*g, m=20 kg and g=9.8m/s^2.

Force is directly translated to acceleration through Newton's 2nd Law, F=ma, so F=ma=mg => a=g. Note that F positive means the object is accelerating towards the ground; I just chose that convention with the previous equation. However, this choice requires us to be consistent in the future. In particular, if we say our object's position is x=10m, we must mean 10m below its starting position at the top of the cliff, since positive means "towards the ground".

Just like distance = rate * time, we also have velocity = acceleration * time from basic Calculus, if acceleration is constant (which it is here). We need to find the time of collision, which will then give us the velocity of the object at that time. We know the object will have moved a distance of d=100m at the time of collision, which would allow us to use the kinematic equations for constant acceleration (listed here (http://wiki.answers.com/Q/What_are_the_kinematic_equations)). Again, as it turns out, they can all be derived from basic Calculus. The most directly useful equation is probably

v(t)^2 = v(0)^2 + 2a*(x(t)-x(0))

where I've made v and x explicitly functions of time. The starting position of the object is at height 0m, and the ending position it at height 100m. So, x(0) = 0m, x(c) = 100m, where c means the time of collision with the ground. Plugging in, we have

v(c)^2 = 0^2 + 2*9.8m/s^2*(100m-0m) = 1960 m^2/s^2
=> v(c) ~= 44.27 m/s.

The final kinetic energy must of course be 1/2*m*v(c)^2 = 1/2*20 kg*1960 m^2/s^2 = 19600 J


The second way is much faster in this case (and, as it turns out, "most" of the time). You require conservation of energy. There are two sources of energy relevant to this problem: kinetic and gravitational potential. Their sum, that is, total energy, must always be constant. Initial kinetic energy is 0 since initial velocity is 0. Initial gravitational potential energy is m*g*h(0) where h(0) gives the initial height above the cliff, in this case, h(0)=100m. Final gravitational potential energy is m*g*h(c), where the height at the time of collision is clearly h(c)=0m. So, we have...

Total energy (0) = KE(0) + GPE(0) = 0 + 20kg*9.8m/s^2*100m = 19600 J.
Total energy (c) = KE(c) + GPE(c) = KE(c) + 20kg*9.8m/s^2*0m = KE(c) = 19600 J.

The final velocity can be found from the final kinetic energy by inverting the standard formula

KE = 1/2 * m * v^2
=> Sqrt(2 KE / m) = v
=> v(c) = Sqrt(2 * 19600 J / 20 kg) ~= 44.27 m/s


A startlingly insightful observation is that in one case we calculated velocity and then found energy, and in the other case we calculated energy and then found velocity. It turns out that Lagrangian Mechanics (http://en.wikipedia.org/wiki/Lagrangian_mechanics) can basically perform the second method on horrendously complicated problems. It "keeps track" of something like energy to calculate the position and velocity of objects. It's a beautiful idea, really; mathematically, it reduces basically all of classical mechanics to solving differential equations. In a (very short) nutshell, classical mechanics becomes the application of the Euler-Lagrange equation (http://en.wikipedia.org/wiki/Euler-Lagrange) to the action integral (http://en.wikipedia.org/wiki/Action_Integral) which then satisfies Hamilton's Principle (http://en.wikipedia.org/wiki/Hamilton's_Principle). Put another way, Hamilton's Principle directly implies all of classical mechanics in a way that allows very mechanical computation.

For this problem, Lagrangian mechanics is wild overkill. Still, for more complicated systems like the double pendulum (http://en.wikipedia.org/wiki/Double_pendulum), this is basically the only sane way to go.