How is kinetic energy affected by different reference frames?

[penutbutr]

Homework Statement

Calculate the energy needed for 1 kg mass at the velocity of 2 m/s go up to the velocity of 6 m/s.

Homework Equations

E_k = mv²/2

The Attempt at a Solution

From the mass point of view, it is stationary and goes up to a velocity of delta (6-2) m/s = 4 m/s
so the energy needed must be E_k = 1*4²/2 J = 16/2 J = 8 Joules

But the answer is 16 Joules.
E_before = 1*2²/2 = 4/2 Joule = 2 Joule
E_after = 1*6²/2 = 36/2 Joule = 18 Joule
E_delta= E_after-E_before = (18-2) Joules = 16 Joules

How can the answer not be 8 Joules? From the mass perspective, it is stationary and just goes up 4 m/s.

 

[BvU]

Hello penutbutr,

With your reasoning a succession of accelerations with only 1 m/s would be energetically cheaper still, so there must be something wrong. Can you fathom what ?

 

[penutbutr]

So in order to get correct calculations. "v" in the formula mv²/2 can only be measured by someone in rest relative to the mass?

 

[penutbutr]

How can it be harder to accelerate an extra 1 m/s, in high speed like 50 m/s ? than it would be in low speed like 2 m/s. That is what the latter method implies, right?

 

[BvU]

The speed and the kinetic energy should be in the same frame of reference. In your first post

From the mass point of view, it is stationary and goes up to a velocity...

the second part isn't correct: from the mass point of view it is stationary and stays stationary...

 

[penutbutr]

The speed and the kinetic energy should be in the same frame of reference. In your first post
the second part isn't correct: from the mass point of view it is stationary and stays stationary...

Oh, That makes sense!
So kinetic energy is always relative to something, the more speed an object has the more destruction it can cause at the object in rest. Meanwhile an object traveling close to the speed of the other object will perceive it as having less kinetic energy.

 

[BvU]

If you are familiar with the SUVAT equations for uniform acceleration:

If you push with a force of 1N on a mass of 1 kg for 2 seconds, it speed is 2 m/s $\ \$ ($at$), $\ \$ and you have pushed over 2 m $\ \$ (${1\over 2}at^2$). $\ \$ The work is 2 J and the kinetic energy is 2 J as well.

Now push with the same force for another 2 seconds. The speed increases to 4 m/s and you have pushed over $\ \ 2*2 + {1\over 2}*1*2^2 = 6$ m. $\ \$ The work is 6 J and the kinetic energy has increased by 6 J from 2 J to 8 J, which is equal to ${1\over 2}mv^2$​

 

[CWatters]

I wouldn't use SUVAT to solve the OP. Just stick with calculating the change in energy.

 

[BvU]

I wouldn't use SUVAT to solve the OP. Just stick with calculating the change in energy.

I don't think solving is the subject. Thread starter did that. Understanding the reference frame for kinetic energy and speed is the issue.

 

[penutbutr]

If you are familiar with the SUVAT equations for uniform acceleration:

If you push with a force of 1N on a mass of 1 kg for 2 seconds, it speed is 2 m/s $\ \$ ($at$), $\ \$ and you have pushed over 2 m $\ \$ (${1\over 2}at^2$). $\ \$ The work is 2 J and the kinetic energy is 2 J as well.

Now push with the same force for another 2 seconds. The speed increases to 4 m/s and you have pushed over $\ \ 2*2 + {1\over 2}*1*2^2 = 6$ m. $\ \$ The work is 6 J and the kinetic energy has increased by 6 J from 2 J to 8 J, which is equal to ${1\over 2}mv^2$​

Did this method now, and got the answer 18 joule which verifies it

 

[haruspex]

So in order to get correct calculations. "v" in the formula mv²/2 can only be measured by someone in rest relative to the mass?

Yes. KE is not the same in different inertial frames. Consider a reference frame moving up at 4m/s. In this frame there is no change in KE.

got the answer 18 joule

You mean 16J, right?