The relationship between potential energy and kinetic energy

[ThievingSix]

In class it was explained that at half way between a given distance Potential energy(PE) lost equals the Kinetic energy(KE) gained. OR

PE Lost = KE Gained at 1/2 the distance.

But when i do the math for it, it does not work out. What am i missing, i understand the concept but i can't prove it? And in theory it sounds correct because of the law of conservation of energy.

Here is the situation(i made it up to prove that the statement is correct).

A stationary mass is dropped from a distance of 1m and the only force that acts upon it is the Earth's gravity, it hits the ground in 1 second and its motion is along the Y axis only. Calculate PE and KE at "X".

[PLAIN]http://img854.imageshack.us/img854/3959/diagram.png

This is what i have so far,

Distance = 1m
Time = 1s
Mass = 10kg
Acceleration = 9.8 m/s/s
Initial speed = 0 m/s
Final speed = 9.8 m/s
Force = 98N(toward's the center of the earth)
Gravitational Potential energy(mass*gravity*height) = 10*9.1*1 = 98 kgms
Potential energy at rest(mass*acceleration*displacement) = 10*9.8*1 = 98 kgms
Kinetic energy at rest(1/2 Mv²) = 1/2*10*0 = 0 kgms

This is where it gets confusing,

PE at "X" = 10*9.8*0.5(half the displacement) = 49 kgms(which should be correct)
KE at "X" = 1/2*10*4.9²(velocity at "X") = 120.05 kgms

Therefore PE lost \neq KE gained at half the distance?

I've done something wrong, but what?, I've ran through this so many times i can't even see my own error.

 

[Vagn]

Your velocity at point X is wrong, it should be the square root of g. You need to solve the UVAST equations for x=0.5 and then use the time taken to reach X to find your velocity at point X.

 

[douglis]

KE at "X" = 1/2*10*4.9²(velocity at "X") = 120.05 kgms

Your velocity is wrong.You probably assumed that the time it took the mass to reach the x point was 0.5sec which is not right.

 

[UnsungHero]

There are several problems in the problem you have published. Firstly, "A stationary mass is dropped from a distance of 1m and the only force that acts upon it is the Earth's gravity, it hits the ground in 1 second and its motion is along the Y axis only. Calculate PE and KE at "X". " is wrong, because it wil not take a body 1 sec to fall 1 m.

Galilei's law of the free fall state, that s = 1/2*g*t^2. So in 1 sec the distance is s = ½*9,82*(1)^2 = 4,91m

You have to use this law to calculate the time used for falling s meters. (s =½ m )

s= ½*g*t^2. t => t= sqrt(2s/g) and then you have to insert this in v = g*t = g*sqrt(2s/g)=sqrt(2s*g), and you get, that the kinetic energi (and thereby the change in kinetic energy) is
Ekin= ½*m*v^2 =½*m*(2s*g) = m*s*g and that is exactly the change in potential energy by falling s meters.

The only numbers you have to know - to do your calculations - is s and m.

But you do not have to know the mass to prove your statement. Just use Galileis law for falling bodies, which comes out of Newton's second law F = m*a

 

[sophiecentaur]

Your velocity at point X is wrong, it should be the square root of g. You need to solve the UVAST equations for x=0.5 and then use the time taken to reach X to find your velocity at point X.

Yes - you always need to choose the right SUVAT equation to get what you need ONLY from what you have. You must not make 'assumptions' about values unless you have calculated. This is one of the most common problems for students starting on Motion problems.

 

[ThievingSix]

Alright this is how i got the answer, please tell me if its the right way.

I used the method Vagn suggested.

Finding velocity at "X"
v² = u² - (2gs)
v² = 0² - (2*9.8*0.5)
v² = -9.8

v = \sqrt{9.8}

Finding KE at "X"

KE = \frac{1}{2}\times m \times v²

KE = \frac{1}{2}\times 10 \times (\sqrt{9.8})²

KE = 49 kgms

Therefore PE lost = KE gained at X

 

[ed8widge]

proper:
since you are taking it at the half way ,
D=0.5
m=10
t= 1s and half time is 1/2 s
Acc=g=9.8
final speed=the speed half way =Vf= √9.8 if time is half
PE=mgh
Ke=1/2mV²
Pe=10x10x0.5=50
Ke=1/2x10x(√9.8)²=49.92

 

[the_emi_guy]

Alright this is how i got the answer, please tell me if its the right way.

I used the method Vagn suggested.

Finding velocity at "X"
v² = u² - (2gs)
v² = 0² - (2*9.8*0.5)
v² = -9.8

v = \sqrt{9.8}

ThievingSix

Do you know where these "UVAST" equations & Galilei's law come from and why your initial, simpler, assumption is not valid?