Does Accelerating at Higher Speeds Require More Energy?

[_Enigma_]

OK. I have a problem. I've just started taking physics (1 month) so for more experienced physicists this should be simple.
The kinetic energy of an object is: KE = 1/2*m*v^2
Right?

This means that a spaceship would use a lot more energy to accelerate from 100m/s to 105m/s than it would do by accelerating from 0m/s to 5m/s.
But this is contrary to what my intuition tells me. My intuition tells me that it uses the same amount of energy in both cases. Am I wrong?

Some examples that support my idea:

• 
  According to general relativity, everybody has equal right to claim that they are standing still, and that everything else is moving.
  So, let's say that space shuttle A is standing still. Right next to it is space shuttle B which is also standing still. On the opposite side of Space shuttle B, is space C. Space shuttle C is traveling at 100m/s. Then space shuttle B accelerates to 5m/s in the opposite direction of the motion of space shuttle C using a certain amount of energy.
  Now to space shuttle A it's obvious that the energy space shuttle B used is:
  1/2*m*5^2 - 1/2*m*0^2 = 12.5*m
  But to space shuttle C, which can claim that it's standing still and that Space shuttle A is traveling at 100m/s, whereas Space shuttle B was traveling at 100m/s alongside A, but is now traveling at 105m/s it seems that the energy used to accelerate space shuttle B was:
  1/2*m*105^2 - 1/2*m*100^2 = 512.5*m, which is a lot more than what space shuttle A thinks that space shuttle B used.
  Obviously there is something wrong here. I mean, space shuttle B used only a one amount of energy. So what's wrong?

• 
  A man standing still on a train traveling at 100m/s accelerates from 0m/s to 5m/s. To do this he used a small amount of energy. But to a person standing still outside the train it seems that he used a lot more energy just like in the example above

• 
  An object falling freely (neglecting air resistance) will accelerate at a constant rate. But the force exerted on it by gravity is also constant. This doesn't fit with KE = 1/2*m*v^2

Please help me! What is wrong? Oh, and I'm not very good at maths (2nd year high school) so try to make the explanation simpel.
thx

 

[Tide]

Your problem is that you are changing reference frames between the two calculations.

In a stationary frame of reference the work required to accelerate an object (say A) from 0 to 5 m/s is less than the energy required to accelerate an object from 100 to 105 m/s (say B). However, as viewed from a reference frame moving at 100 m/s relative to the first reference frame, the work required to accelerate object B from 0 to 5 m/s is less than the energy required to accelerate object A from -100 m/s to -95 m/s.

That simply means that kinetic energy depends on speed and relative to any frame of reference a stationary object has no kinetic energy while a moving object does.

 

[Physics is Phun]

I think you may be getting force and energy mixed. It would take the same force in both cases. but not the same energy. think of riding a bike. it takes little effort(energy) to go from 0 to 5km/h but it is significantly harder to go from 35 to 40 km/h

 

[aekanshchumber]

You are mixing whatever you have learned, mixing force with energy, mixing frames of references and everything. it seems that you even do not pay attention in your classes, k.E = 0.5mv^2 tells us about the energy present in a body due to the motion of the body.
By increasing the speed you will increase the K.E, in bote the cases increase in the K.E is same.
Also, please do not pay much attention to the R. Theory at this stage specially. Keep frame of reference in mind only, try to understand the concepts, hope these thing might help you in great deal.

 

[pervect]

OK. I have a problem. I've just started taking physics (1 month) so for more experienced physicists this should be simple.
The kinetic energy of an object is: KE = 1/2*m*v^2
Right?

This means that a spaceship would use a lot more energy to accelerate from 100m/s to 105m/s than it would do by accelerating from 0m/s to 5m/s.
But this is contrary to what my intuition tells me. My intuition tells me that it uses the same amount of energy in both cases. Am I wrong?

Yes, your intuition is wrong, and the equation is correct.

Some examples that support my idea:

According to general relativity, everybody has equal right to claim that they are standing still, and that everything else is moving.

Just thank pitchforks & pointed ears that you are not dealing with energy in general relativity. (We've got a few threads going on this in the relativity forum. It is considerably more complicated than the problem you are dealing with.) As others have advised, stick to a consistent reference frame, and leave considerations of general relativity till later, when you are learning general relativity.

<snip>

A man standing still on a train traveling at 100m/s accelerates from 0m/s to 5m/s. To do this he used a small amount of energy. But to a person standing still outside the train it seems that he used a lot more energy just like in the example above

This is correct, though it may not be intiutive. In the reference frame of the man standing on the train, it takes a small amount of energy to accelerate from 0 to 5 m/s. To a person standing outside the train, it takes more energy.

Energy is still conserved, though. If the train maintains a constant velocity, the simplest case, the train maintains its velocity only because the train's engine does the needed extra work.

An object falling freely (neglecting air resistance) will accelerate at a constant rate. But the force exerted on it by gravity is also constant. This doesn't fit with KE = 1/2*m*v^2

This should be the easiest problem of them all, and be the easiest to see why the equation for energy is correct. In a uniform gravity field, the potential energy is m*g*h

The sum of the kinetic energy and the potential energy is constant

i.e

$$m g h + \frac{1}{2} m v^2 =E$$

where E is constant.

You can use this equation to find the velocity of an object that starts at rest after it has dropped h meters. Initially, h=h0 and v=0. When the object hits the ground, h=0 and v = vf.

m * g * h_0 + 0 = 0 + .5 * m * vf^2

thus

$$vf = \sqrt{2 g h}$$

You can also get this equation from the equation of uniform accelerated motion

$$h = \frac{1}{2} g t^2 \hspace{.5 in} t = \sqrt{\frac{2 h}{g}} \hspace{.5 in} v = g t = g \sqrt{\frac{2 h}{g}} = \sqrt{2 g h}$$