Consider a stationary 10kg bike; if I apply a force of 10J then $KE=\frac{1}{2}mv^2$ so $v=\sqrt\frac{2KE}{m}=1m/s$. If we consider the same example again, but from the point of view of a stationary astronaut looking down from space, the Earth is moving at approximately 30,000m/s, so the KE of the stationary bike would be $KE=\frac{1}{2}10*30,000^2=4,500,000,000J$. If we assume the bike is being pushed in the same direction the Earth is travelling, to get to 1m/s the bike will need a kinetic energy of $KE=\frac{1}{2}10*30,001^2 = 4,500,300,005J$, which is 300,005J more than for an observer on Earth. I understand that conservation of kinetic energy only makes sense if you look at all objects from the same frame, so my question is why it would appear to take so much more energy to accelerate a bike by 1m/s from one observer's point of view than another?
34385
The energy of a moving object is $E = mv^2\;.$ That is it increases with velocity squared.  I walk at say 3 miles per hour, or lets round that down to 1 meter per second for a slow walk. I weigh less than $100~\mathrm{kg}\;,$ but lets just round that up to $100~\mathrm{ kg}$ for convenience (it is just after Christmas).  So when I walk along the pavement, I have $100~\mathrm{kg\; m^2 s^{-2}}$, 100 joules of kinetic energy.  Now I get on a passenger jet, which is cruising at around 500 knots, call that 250 meters per second.  In my seat I have $100\times 250^2 = 6250000$ joules of kinetic energy. But when I walk down the aisle I have $100\times 251^2 = 6300100$ joules of kinetic energy. The difference between these is: 50100 joules.  It feels the same to me, walking down the pavement as walking down the aisle of the plane. I didn't have to make some huge effort to get up to speed on the plane, yet I needed 500 times the energy to do it.  How is this possible, and where did the energy come from?
61869
I thought this would be a particularly simple problem but it is turning out to be quite the opposite. I am sure I am doing a very simple mistake.   The problem statement is that there is a mass which is just barely kept on the spring (help by the force equal to the weight of the spring) such that the spring is uncompressed. As soon as I let the object go, it will compress the spring and come to rest at some height. The energy stored in the spring will be equal to the difference in the potential energy at the two height of the spring (compressed and uncompressed),  $$mgh_i -mgh_f= 1/2 kx^2$$ Furthermore, the compression of the spring would just be the difference in the $h_f -h_i$. This gives me two roots for compression,  $$ x=0 $$ And the second would be, $$ mg=1/2kx $$ But then by this, $kx=2mg$ and $kx$ is force by Hooks law, but then does it means that the force compressing the spring is twice the weight of the object? That sounds odd. I was expecting I would just recover Hooks law but I guess I am doing something wrong here. But I am not sure what that is.   Any help will be much appreciated.  P.S. This is not a homework problem. We are designing a project for our school.  Thanks for your time.
63588
Here is a classic problem:     A 60kg boy and a 40kg girl are sitting in front of each other on skateboards on a frictionless horizontal surface. The boy pushes the girl and the girl gains a velocity of 0.3m/s during the 0.5s the boys hand was in contact with her. What is the boy's resulting velocity?   Considering Newton's Third Law, the best way is to find the total force applied and rearrange the formula to find the boy's speed.  Newton's Law Method:  $F = m*a$  $a=\frac{0.3m/s}{0.5s}$  $F = 40kg * 0.6m/s^2$  $F = 24N$  Therefore the boy's speed will be:   $\frac{24N}{60kg} = a$  $a = 0.4m/s^2$  $v = 0.4m/s^2 * 0.5s$  $v = 0.2m/s$  Kinetic Energy Method:  $E_{k}=\frac{1}{2}*m*v^2$ the girl's kinetic energy  $E_{k}=\frac{1}{2}*40kg*(0.3m/s)^2$  $E_{k}=1.8J$  Since every action has an equal and opposite reaction the boy should have the same amount of kinetic energy.  $E_{k}=\frac{1}{2}*60kg*(0.2m/s)^2$  $E_{k}=1.2J$  As you can see, using Newton's Third Law, the kinetic energy was not the same. So what's going on here? I thought that in this case they should both have the same amount of kinetic energy due to Newton's Law. I must be missing something since there is more energy going in one direction than the other after meeting. My book used the first method as the answer and, well, it feels wrong.
82924
How much time would it take for a traveler (in traveler's perspective) to reach a star at distance $d$, if it accelerates at 9.8 m/s² (for a comfortable travel)?  I understand that $V = V_0 + a\times t$ and such related high-school physics, and I know there is something called Lorentz transformations that I can grab the formulae from the Internet, but I can't really put the numbers together to get what I want, so I would also like to know how to derive the answer to the question from the basic equations.  Also, it is not quite clear to me the role of the $E = mc^2$ in this travel, specifically: in travellers perspective, can it accelerates at 9.8 m/s² indefinitely, using always a constant energy output? In other words, would the ship's engine "feel" the ship heavier, or the mass increase effect would only be observable from another frame of reference?
45829
Work can be calculated by $W = Fd\cos\theta$ where $\theta$ is the angle between the force $F$ and the displacement $d$  Let's say there's a ball and a (physically ideal - no friction etc.) robotic arm situated in otherwise empty space. The arm takes the ball, moves it around in a circle, and then returns it exactly to where it started. The arm also returns to its starting position.  In this case, there is no displacement overall. The ball and the robot arm are in the exact same positions as when they began.  Thus, $W = F\cdot 0 \cdot \cos\theta = 0$ So no energy was required to move the ball in a circle.  However, this disagrees with my intuition, because if I were to make a robot arm that would do this, I feel like I would need to give it an energy source (for example, a battery), and that by the time the robot was done, I would have lost energy from that battery.  Does the movement require energy? If so, where did I mess up in my calculation? If not, why would my robot need a battery pack or some energy source in order to make this movement? (If it doesn't, please explain.)
120269
Suppose we have an object in space. say gravity is negligible, and the thing has a rocket on both sides. the thing fires one rocket for, say, 5 seconds. now it's moving. The object has kinetic energy. Now the other rocket fires for 5 seconds. Then it's stationary. where does the energy go? I know that some of the energy from the rockets would just go into space as heat and kinetic energy of the hydrogen particles flying out, but how does there seem to be more energy in the situation if the first rocket just goes for 10 seconds than if one goes for 5 and the other goes for 5?
8542
In (non-relativistic) classical physics, if the temperature of an object is proportional to the average kinetic energy ${1 \over 2} m\overline {v^{2}}$of its particles (or molecules), then shouldn't that temperature depend on the frame of reference - since $\overline {v^{2}}$ will be different in different frames?  (I.e. In the lab frame  $K_l = {1 \over 2} m\overline {v^{2}} $, but in a frame moving with velocity $u$ relative to the lab frame, $K_u =  {1 \over 2} m \overline {(v+u)^{2}}$).
121121
If it takes energy to slow an object down, and then the object also loses KE, then how is energy conserved? Don't you have a net loss of energy?
95274
I believed I could use $0.5\times M_{ej} V_{ej}^2$, with $M_{ej}$ being the ejected mass and $V_{ej}$ being the velocity of the ejected mass. But I noticed in  and  that the the mean velocity is somehow calculated as following : $V_{mean}=(3/5)^{0.5}\times V_{ej}$. And so the kinetic energy was calculated as : $(3/5)(0.5)\times M_{ej} V_{mean}^2 =(3/10)\times M_{ej} V_{mean}^2$. So is there any explanation for how this factor of $0.3$ or the mean velocity was calculated ?
83860
I am stuck in a question where energy conservation is failing and momentum conservation is correct. I think I might be doing something wrong, that's why I'm asking this question. The problem is as follows:     A bullet of mass $25\,g$ is fired horizontally into a ballistic pendulum of mass $5.0\,kg$ and gets embedded in it. If the centre of pendulum rises by a distance of $10\,cm$, find the speed of the bullet.   (H.C. Verma, Centre of Mass, Q47)     Let the mass of the bullet be $m=25\,g$. Let the mass of the pendulum be $M=5\,kg$. Let the peak height be $h=0.1\,m$. Let the initial and final velocities be $u$ and $v$ respectively.  Method 1: (Momentum Conservation)  $$mu=(M+m)v \implies v=\frac{mu}{M+m}\\ \text{Also, }\frac{1}{2}(M+m)v^2=(M+m)gh\\ \implies u^2=2\biggl(\frac{M+m}{m}\biggr)^2gh\\ \implies u=201\sqrt2 \text{ m/s}$$  Method 2: (Energy Conservation)  Since both the masses move together after the collision, and because the velocity at highest point is null, therefore: $$\frac{1}{2}mu^2=\frac{1}{2}(M+m)v^2=(M+m)gh\\ \implies u^2=2gh\biggl(\frac{M+m}{m}\biggr) \implies u=\sqrt{402}$$    To summarize myself, I am curious about the following: Why do the results differ? Shouldn't the energy be conserved as well as the momentum? Since it should be, then what is the flaw in my calculations?
121441
Earth's orbit is a slight ellipse, so to conserve momentum its speed increases when it is closest to the Sun. If the speed changes there is an acceleration. If there is an acceleration there is a force. Even if the change is small and gradual, wouldn't we experience a force because the Earth is so massive?
713
I was watching a youtube video the other day where an economist said that he challenged his physics professor on this question back when he was in school. His professor said each scenario is the same, while he said that they are different, and he said he supplied a proof showing otherwise.  He didn't say whether or not the cars are the same mass, but I assumed they were. To state it more clearly, in the first instance each car is traveling at 50mph in the opposite direction and they collide with each other. In the second scenario, a car travels at 100 mph and crashes into a brick wall. Which one is "worse"?  When I first heard it, I thought, "of course they're the same!" But then I took a step back and thought about it again. It seems like in the first scenario the total energy of the system is the KE of the two cars, or $\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$. In the second scenario, it's the KE of the car plus wall, which is $\frac{1}{2}m(2v)^2 + 0 =  2mv^2$. So the car crashing into the wall has to absorb (and dissipate via heat) twice as much energy, so crashing into the wall is in fact worse.  Is this correct?  To clarify, I'm not concerned with the difference between a wall and a car, and I don't think that's what the question is getting at. Imagine instead that in the second scenario, a car is crashing at 100mph into the same car sitting there at 0mph (with it's brakes on of course). First scenario is the same, two of the same cars going 50mph in opposite directions collide. Are those two situations identical?  PS: This scenario is also covered in an  of .
100737
Why does the kinetic energy of a particle moving in circular motion increase when the turn radius decreases and there is no torque acting? E.g. if a planet is rotating about its axis and it shrinks to half of its radius then according to conservation of angular momentum velocity will doubled, i.e. its kinetic energy will increase. Does this mean that without any external force the kinetic energy of any thing can be increased? Please explain!
103546
So I have some confusion here, I am sure I knew this at some point.  Let's say an object of 10 kg is dropped from a height of 10 m.  When it reaches the ground, the work done on the object should be the force ($mg$) x distance or 10 kg x 9.8 m/s/s x 10 m.  That gives 980 joules of work done on the object by gravity.  But the object did not gain 980 joules of .  It lost 980 joules of GPE and gained 980 joules of kinetic energy (up to the point of it reaching ground level).   Using the change in GPE and KE, it looks like no work was done on the object because the loss in GPE equals the gain in KE.  So   a) am I right that no net work was done on the car by sum of all forces?    b) is the work done by gravity equal to force time distance, or is it equal to the change in mechanical energy of the object which is zero?
57534
I read in  in this site that the KE a free-falling ball acquires is not originated by the attracting body but that energy was actually stored in the ball when it had been lifted to the height it dropped from.  In this way, it was said, gravity is subject to the conservation of energy principle and cannot change the total energy of an object.  Consider now the maneuver known as a  (also gravity assist) used by space probes such as the . A space probe approaches a planet with velocity $v$, slingshots around, and ends up with velocity $v+2U$, where $U$ is the velocity of the planet.  Consider the energy of the probe. Before, it was $E_i=\frac{1}{2} mv^2$ and after it is $E_f=\frac{1}{2} m(v+2U)^2$. It looks like $E_f$ is much bigger than $E_i$ - but where did the additional energy come from?   Is this not a violation of the conservation of energy principle?
35120
Two twin sisters synchronize their watches and simultaneously (from the earth frame) depart earth in different directions. Following a predetermined flight plan, each sister accelerates identically to 99.9%c and then returns home at the same time (again in the earth frame). The observers on earth see each twin as having aged identically, as the symmetry of the problem dictates. Each twin however should see the other as having aged... What gives?
42378
If the velocity is a relative quantity, will it make inconsistent equations when applying it to the conservation of energy equations?  For example:  In the train moving at $V$ relative to ground, there is an object moving at $v$ relative to the frame in the same direction the frame moves. Observer on the ground calculates the object kinetic energy as $\frac{1}{2}m(v+V)^2$. However, another observer on the frame calculates the energy as $\frac{1}{2}mv^2$. When each of these equations is plugged into the conservation of energy, they will result in 2 distinct results (I think).