# Question about work and energy transfer

Homework Helper
Something is bugging me... Is there something wrong with thinking that if an object A exerts a constant force F on object B, through a distance d, then A transfers F*d joules of energy to B?

Consider a situation in space... an astronaut mass m, and a rock m (no other objects anywhere). The astronaut exerts a force F on the rock (let's say along the x axis) for some time t. Let's say the mass acquires a velocity v, so the astronaut acquires velocity -v.

If I were asked, how much chemical energy does the astronaut lose... The answer seems to me to be: mv^2 (sum of the final kinetic energies of both the astronaut and the mass). There's no other source for the kinetic energy to come from.

But if I were to use this idea of energy transfer (force through distance = energy transferred) then, I'd get the result that the astronaut transferred 1/2(mv^2) to the rock (no problem), and the rock transferred 1/2(mv^2) to the astronaut (big problem here!)...

Last edited:

Is it really a problem? the astroid exerted a normal force on astronaut equal to that of the force applied by the astronaut. Both have the resulting speeds of v, with opposite velocities. In this case the astronaut provided all the energy for the colision. Does it violate any laws of physics?

Or is there something wrong with what you said?

Idk, Just an opinion. Im not very smart.

Last edited:
Homework Helper
Yapper said:
Is it really a problem? the astroid exerted a normal force on astronaut equal to that of the force applied by the astronaut. Both have the resulting speeds of v, with opposite velocities. In this case the astronaut provided all the energy for the colision. Does it violate any laws of physics?

Or is there something wrong with what you said?

But thinking in terms of work done by the astroid on the astronaut, and work done by the astronaut on the astroid it seems like... the astronaut transfers (1/2)mv^2 to the asteroid, and the asteroid transfers (1/2)mv^2 to the astronaut.... Is this consistent with all the energy coming from the astronaut? If it is, how exactly...

Q_Goest
Homework Helper
Gold Member
The astronaut and rock can be viewed as a single mass, instead of two different masses. Both astronaut and rock increased their kinetic energy, both had a force on them. The equation would be
Et = Ft*d = 1/2 Mt*v^2
where Et = Total energy used to accelerate both astronaut and rock
Ft = Total force produced by the rocket used to accelerate the astronaut and rock
Mt = Total mass, the sum of the astronaut and rock masses
v = velocity

You could look at it another way: the rock had a force on it, resulting in energy
Er = Fr*d = 1/2 * Mr*v^2
where Er is the energy of the rock
Fr = the Force on the rock
d = distance the force acts through
Mr = Mass of the rock
v = velocity

The astronaut had a much larger force on him, along with a force from the rock. So the astronaut's energy is
Ea = (Ft - Fr)*d = 1/2 * Ma*v^2
where Ea = Energy of the astronaut
Ma = mass of the astronaut.
(Note: Ft = Fa + Fr where Fa = force on astronaut)

Now the first equation should equal the sum of the second two:
Et (or) Ft*d = Fr*d + Fa*d

or looking only at kinetic energy, the first should equal the sum of the second two:
1/2 Mt*v^2 = 1/2 * Mr*v^2 + 1/2 * Ma*v^2

It all should work out, regardless of which perspective you choose.

Homework Helper
Q_Goest said:
The astronaut and rock can be viewed as a single mass, instead of two different masses. Both astronaut and rock increased their kinetic energy, both had a force on them. The equation would be
Et = Ft*d = 1/2 Mt*v^2
where Et = Total energy used to accelerate both astronaut and rock
Ft = Total force produced by the rocket used to accelerate the astronaut and rock

There's no rocket. The astronaut pushes against the rock, so he and the rock go in opposite directions. Force on the rock is equal and opposite to the force on the astronaut.

Q_Goest
Homework Helper
Gold Member
oops! sorry... Choose a reference frame which isn't accelerating with respect to either the rock or astronaut. Use this reference frame to determine through which distance the force is applied. In other words, the rock and the astronaut move through a distance d with respect to a 'stationary' reference frame... the reference frame they were in before the astronaut started pushing.

Astronaut now has energy:
Ea = F * da = 1/2 Ma*Va^2
where da = distance astronaut moves through
Va = velocity of astronaut

Rock has energy:
Er = F * dr = 1/2 Mr*Vr^2
where dr = distance rock moves through
Vr = velocity of rock.

The total energy is Et = Ea + Er.

Note that if the masses are equal, Va = Vr and da = dr

Does that help at all, or is there something else I'm confusing you with???

krab
learningphysics said:
Something is bugging me... Is there something wrong with thinking that if an object A exerts a constant force F on object B, through a distance d, then A transfers F*d joules of energy to B?
Yes, there is something wrong. There is F*d work done, but that doesn't mean that the work is done specifically ON B. This amount of energy just gets distributed according to kinematic laws like conservation of momentum. So the atronaut gets some, and the rock gets some.

To make your example simpler yet, consider that there are two rocks, each with a spring mechanism that exerts your force F. In the first case, each lets its spring go at the same time, and these act to accelerate the rocks apart. In the second instance, only one rock acts. Now assume that in both cases, the final velocities are the same. You find by analyzing the situation that in the first case, each spring acted through d/2 and each did Fd/2 work. In the second case, where you would say only rock A did work on B, you find that A's spring acted through distance d. Same amount of work done, same final result. It doesn't matter who the spring belonged to.

Homework Helper
krab said:
Yes, there is something wrong. There is F*d work done, but that doesn't mean that the work is done specifically ON B. This amount of energy just gets distributed according to kinematic laws like conservation of momentum. So the atronaut gets some, and the rock gets some.

But the force is acting only on B, and it is the only force that acts on B... Shouldn't any redistribution of energy be caused by more work (more forces through distances)...

krab said:
To make your example simpler yet, consider that there are two rocks, each with a spring mechanism that exerts your force F. In the first case, each lets its spring go at the same time, and these act to accelerate the rocks apart. In the second instance, only one rock acts. Now assume that in both cases, the final velocities are the same. You find by analyzing the situation that in the first case, each spring acted through d/2 and each did Fd/2 work. In the second case, where you would say only rock A did work on B, you find that A's spring acted through distance d. Same amount of work done, same final result. It doesn't matter who the spring belonged to.

In your spring mechanism example, there is an equal reactional force, even when only one spring acted. I'd think I'd still be able to say: rock A exerted a force F on rock B, and by newton's third law, rock B exerted an equal force F on rock A over the same time... And since the masses are equal, the accelerations are equal and the distances the forces act are equal. So rock A did Fd work on rock B, and rock B did Fd work on rock A.

Thanks to everyone who has responded so far!

Homework Helper

Ok the atronaut produces a force F which moves both him and the rock away from the point of origin at velocity v.

Conservation of energy plays a role here but so does conservation of momentum. Newtons third law for every action there is an equal and opposite reaction. Therefore when the atronaut pushed on the rock producing movement of the rock away from the origin at v the astronaut also leaves at v. This all stems from an idea that momentum is always conserved, p=mv there is no net momentum in the system so the two momentums have to equal each other p1+p2=0 p1=p2 mv+m(-v)=0
The energy he supplies is cut in half by the two systems so to speak, if you want to look at it this way.
Another situation that may arise is an astronaut of 100kg is trapped away from his shuttle 10m the only thing he has is a 1kg wrench in order to return he throws the wrench at 10m/s 180 degrees from the direction of the spaceship.
Because of conservation of momentum the wrench ends up with 10kgm/s of momentum as does the astronaut.... the astronaut moces towards the ship at .1m/s and takes 100 seconds to return to the ship all the energy supplied into the system came from the astronaut . It is the same case with the meteor the total energy supplied can be calculated like this 1/2(mtotal)(vtotal)^2 i hope this all makes sense let me know if it does or not.

Homework Helper
Thanks to BigStelly and everyone else that responded. I believe my major error was using work=Fd with d as distance moves by the center of mass instead of d as the distance traveled at the point of application of the force.

I am facing some troubles too :(

difficulties in law of conservation of energy:
hello,I am from bangladesh and i have completed my a levels a year ago.I got 90 percent in physics and i also teach physics and love it like hell.Here is a problem that has been haunting me to death :(

In this hypothetical situation the universe consists of just two masses,one massive similar to the earth and the other one similar to the moon.The earth here is stationary at the centre of the universe.The moon is moving towards the earth from a very far distance(so far tht the earths gravitational field has no influence over it).As the
moon is moving it for sure posseses kinetic energy.Slowly the moon approaches the earth's gravitational field and accelearates towards the earth and the earth too accelarates towards the moon.The moon's kintetic energy here is increasing but what energy is being converted to kinteic energy.Most people would say gravitational potential energy is being converted to kinetic energy.ok.but which type of energy was initially converted to potential energy which is later converted to kintetic energy.to me it seems tht the earth's gravitational fiels is able to create energy.PLEASE HELP AND ENLIGHTEN ME

Well, nice selection of environment, a completely isolated environment really helps us in concentrating in particular problems.
Stored chemical energy is being converted to kinetic energy which is divided equally to the astronaut and the rock.You might now ask why is energy distributed equally,why this ratio? You are aware of newtons third law of motion,every action has an equal and opposite reaction.
Some other features of the third law pair forces are:
(1)They take place at the same time and for the same time
(2)Both lie in the same line of action
As the astronaut pushed the rock the rock too pushed him with a force of equal magnitude.As both of them are of the same mass and experienced the same magnitude of force,for the same period of time,both accelerated from rest with the same magnitude for the same time.This tells us both of them travelled the same distance.As work done is F*d we now know that work done on both of them is the same.
The astronaut does Y joule of work on the rock and the rock does Y joule of work on the astronaut.For the rock to do work it must have energy,the energy is provide by the astronaut.It seems to be very clear now that the source of kinetic energy of both the astronaut and the rock is the astronauts reserve of chemical energy.
Astronaut lost mv^2 joules of energy to the rock,and half of this energy(1/2mv^2) to the astronaut and possesed the other half.
If someone finds anything wrong with my explanation, please do not hesitate to criticise.

Please if you dnt understand my explanation do not hesitate to criticise ill definately try again

Last edited:
Newtons laws are not that powerful that they can describe where the source of the energy is.Another analogy is a shell being fired from a cannon-the shell goes one way the cannon recoils the other way but their energy comes from the chemical energy of the explosive mixture.