# Apparent energy loss due to acceleration [duplicate]

Imagine two objects of mass $m$ each (both of them in outer space), with initial velocities $v_0$ and $0$. Now I apply a constant force $F$ on both of them for time $t$. Now the change in energies of the masses are $$\Delta E_1=m v_0\Delta v + \frac{1}{2}m\Delta v^2$$ $$\Delta E_2=\frac{1}{2}m\Delta v^2$$ What these equations are saying is that I lose more energy when I try to accelerate an already moving object (same amount of momentum transfer doesn't mean same amount of energy transfer). This doesn't make sense. If I look at it mathematically, i.e $dW = \vec{F} \cdot \vec{ds}$, I completely understand why I'll have to spend more energy on the moving object. But for some reason my intuition says if I apply same force on objects moving at different velocity for the same amount of time, I'll lose the same amount of calories. Where is my intuition wrong?

• This is a super common question! Here is one of many very similar questions, does that answer it for you? Oct 10 '17 at 18:16
• This is because nature behaves in such way,simple. Oct 10 '17 at 18:17
• One thing I have noticed: never rely on your own body mechanics to make intuitive sense of simple physics. Our bodies do so many incredibly complex dynamic operations while they function, that it's really hard to map them back down to simple static situations like this. Oct 10 '17 at 19:06
• Your "intuition" should confirm this if you think about it. Pull a sled or cart with constant force, at a normal, walking speed. Then pull it for the same amount of time while running. If your intuition were correct, why will people run to burn more calories?
– nasu
Oct 10 '17 at 20:23
• Possible duplicate of Where does the extra kinetic energy of the rocket come from? Oct 11 '17 at 12:15

Here's one framing that might make it more intuitive. Suppose you hang from a bar and pull with force $mg$. If your initial velocity is zero, you just sit there and do no work. If your initial velocity is positive, you do a pullup, doing plenty of work in the process. Pullups are harder than hanging.

However in general you shouldn't think about work as 'burning calories' because human muscles behave in weird ways. In this example you would get tired even with zero initial velocity, despite doing exactly zero work. A better example is a weight on a cable: it can hang forever without requiring energy, but it needs a motor to raise the weight up.

Forget about burning calories as you are then dealing with a very complex system (your body) which which convert chemical energy from your food which will not result in a change in the kinetic energy of the ball which you are holding still at arms length ie as far the ball is concerned you are not doing any work on it.

Since the force $F$ is constant the velocity against time graph for your object will be a straight line.

Let the velocity at times $t_1,\,t_2\, t_3$ be $v_1,\,v_2\, v_3$ and $t_3-t_2 = t_2-t_1 = \Delta t$

Since the acceleration is constant then in equal increments of time the increase in velocity is the same $v_3-v_2=v_2-v_1$.

The distance travelled by the object in these equal increments in time is the area under a velocity time graph $s_{12}$ and $s_{23}$.

So the work done by the force in these increments of time is $F\,s_{12}$ and $F \, s_{23}$ and this is equal to the change in kinetic energy of the object in these intervals of time.

Since $s_{12} <s_{23}$ the increase in kinetic energy will be greater the higher the velocity of the object.

So it all boils down to the fact that in a given time $\Delta t$ the impulse on the object $F\Delta t$ changes the velocity of the object by the same amount but the work done by the external force, which is equal to the change in the kinetic energy of the object, is larger in the second interval of time than the first interval of time because in that second interval of time the object has travelled further.

If you were pushing the object and increasing its velocity you certainly would do more work in a given interval of time to increase the kinetic energy of the object but the amount of chemical used by your body would be in excess of the increase in the kinetic energy of the object.