Kinetic/Potential Energy - What am I doing wrong?

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SUMMARY

The discussion centers on the calculations of Kinetic Energy (KE) and Potential Energy (PE) using the formulas PE = mgh and KE = 1/2mv². A user initially calculated the PE of a 10 kg weight lifted 10 meters as 981 Joules but mistakenly calculated the KE as 481 Joules by incorrectly using the final velocity. The correct approach involves applying the SUVAT equations to determine the final velocity, resulting in a KE of 981 Joules, which matches the PE. This demonstrates the conservation of energy principle in physics.

PREREQUISITES
  • Understanding of Kinetic Energy (KE) and Potential Energy (PE) formulas
  • Familiarity with the SUVAT equations for motion
  • Basic knowledge of gravitational acceleration (9.81 m/s²)
  • Ability to perform algebraic manipulations with physical equations
NEXT STEPS
  • Study the derivation and applications of the SUVAT equations
  • Learn about energy conservation principles in physics
  • Explore real-world applications of KE and PE in engineering
  • Investigate the effects of air resistance on energy calculations
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Students studying physics, educators teaching energy concepts, and anyone interested in understanding the principles of motion and energy conservation.

andyg007
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Homework Statement


What am I doing wrong here?

Given the following two formulas for Kinectic Energy (KE) and Potential Energy (PE):

PE = mgh
KE = 1/2mv2

If a 10KG weight was lifted 10 meters above sea level - I calculate its PE as:

PE = 10kg x 9.81m/s x 10meters = 981 Joules

If I let the ball drop I calculate its kinetic energy as (assuming nothing lost due to friction etc):

KE = 1/2 x 10KG x 9.81m/s x 9.81m/s = 481 Joules

However, in this case, I understood the KE should equal the PE?

Which calculation have I got wrong?

Homework Equations


PE = mgh
KE = 1/2mv2

The Attempt at a Solution



PE = 10kg x 9.81m/s x 10meters = 981 Joules

KE = 1/2 x 10KG x 9.81m/s x 9.81m/s = 481 Joules
 
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andyg007 said:
PE = 10kg x 9.81m/s x 10meters
I presume you mean 9.81 m/s2.
andyg007 said:
KE = 1/2 x 10KG x 9.81m/s x 9.81m/s
What makes you think the final velocity is 9.81m/s?
 
I'm thinking this is the amount of time it takes to falls in 1 second and therefore the speed.
 
Take a look at the equations of motion..eg...

V^2=U^2 + 2aS
 
andyg007 said:
I'm thinking this is the amount of time it takes to fall in 1 second and therefore the speed.
You just said that 9.8m/s is the amount of time it takes to fall in 1 second.
Surely the amount of time it takes to fall in 1 second is... 1 second?

Perhaps you mean that it falls 9.8m in the first second, so the average speed over that time is 9.8m/s ... but that is not the case either.
It actually falls 4.9m in the first second.

Have you heard of the kinematic equations (sometimes called suvat equations) and velocity-time diagrams?
Once you have those, you can use the correct one (CWatters tells you which one above) and then do the kinetic energy calculation symbolically.
 
Thanks so much Cwatters/Simon - a great help :-)

-- SUVAT Equations --

I know the following:

(s) displacement = 10 meters
(g) gravity = 9.81 m/s
(t) time = 1 second
(u) initial velocity = 0

I can therefore use:

v2 = u2 + 2as
v2 = 0x2 + 2 x 9.81m/s x 10 meters = 196.2
v2 = 196.2
v = 14.007
----

Back to my original question:

KE = 1/2mv2

KE = 1/2 x 10KG x 14.007 x 14.007 = 981 Joules

This ties up with the PE

:-)
 
andyg007 said:
I know the following:

(s) displacement = 10 meters
(g) gravity = 9.81 m/s
(t) time = 1 second
(u) initial velocity = 0

In the end you didn't use "time = 1 second" to get the right answer but you might like to check if it really does take 1 second to fall 10m.

Perhaps by using another SUVAT equation... v = u+at and solving for t.
 
You did it numerically and showed that it was consistent for one example. maybe that was just a coincidence and you'd get a different result from chosing different numbers? This is why it is better to do the algebra using symbols ...

potential energy lost falling h is ##U=mgh## (close to the surface of a spherical mass with gravitational acceleration at the surface of g)
kinetic energy gained is ##K = \frac{1}{2}mv^2##

from suvat equations, ##v^2=2as## ... in this case, ##a=g## and ##s=h## so:
##K=\frac{1}{2}m(2gh) = mgh = U##

... the advantage of doing it this way is that it now does not matter which numbers you pick, the two equations (related by the laws of motion) are always going to be consistent in this way. notice that it does not even have to be the Earth causing the gravity.
 

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