Mixing
Why is work done on an object greater at higher speeds?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
From this comment by orlp:
If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?
forces work distance
asked 1 hour ago
Pilchard123
1113
New contributor
Pilchard123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
From this comment by orlp:
If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?
forces work distance
asked 1 hour ago
Pilchard123
1113
New contributor
Pilchard123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Net work done on an object such as a rocket always equals its change in kinetic energy.
– David White
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
From this comment by orlp:
If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?
forces work distance
asked 1 hour ago
Pilchard123
1113
New contributor
Pilchard123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
From this comment by orlp:
If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?
forces work distance
forces work distance
asked 1 hour ago
Pilchard123
1113
New contributor
Pilchard123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Pilchard123
1113
New contributor
Pilchard123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Pilchard123
1113
New contributor
Pilchard123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Pilchard123
1113
asked 1 hour ago
Pilchard123
1113
1113
New contributor
Pilchard123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Pilchard123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Pilchard123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Net work done on an object such as a rocket always equals its change in kinetic energy.
– David White
1 hour ago
add a comment |Â
Net work done on an object such as a rocket always equals its change in kinetic energy.
– David White
1 hour ago
Net work done on an object such as a rocket always equals its change in kinetic energy.
– David White
1 hour ago
Net work done on an object such as a rocket always equals its change in kinetic energy.
– David White
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?
To understand this it is useful to consider a “toy model†rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).
Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning†the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.
So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?
Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.
At low speeds most of the fuel’s energy is “wasted†in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.
answered 13 mins ago
Dale
1,039211
add a comment |Â
up vote
2
down vote
There are multiple ways to view this.
The easiest, I think, is that kinetic energy scales with the square of velocity
$$K=frac 12 m v^2$$
If we assume that the rocket booster supplies a constant acceleration, then comparing initial and final velocities we find that
$$Delta v=v_final-v_init=at$$
So for the same amount of time, the change in the velocity is the same, regardless of what the starting velocity actually is. Since we have a squared dependence on the velocity in $K$, this means that the kinetic energy increases more if we started out with a larger velocitiy. i.e.
$$Delta K=frac 12 m v_final^2-frac 12 m v_init^2=frac 12 m (v_final-v_init)(v_final+v_init)=frac 12 m Delta v (Delta v+2v_init)$$
So as we can see, the sum of the velocities in the expression for $Delta K$ is what contributes to a larger change in kinetic energy. Since the work done is equal to the change in kinetic energy, it must be that the rocket does more work when we start off at a larger velocity.
The second way to view this, which you could argue is the same as the first, is to look at the definition of work
$$W=intvec F cdot dvec x$$
Or in one dimension with a constant force
$$W=FDelta x$$
Now, once again assuming a constant acceleration, we know that
$$Delta x = frac 12 a t^2 + v_initt$$
So that the work done is
$$W=Fleft(frac 12 a t^2 + v_inittright)$$
Once again, we see that the initial velocity determines the work. A qualitative explanation from this is that when the velocity is larger, then the object covers more distance in the same amount of time. So if we look at the time the force is being applied, the faster it is moving the larger distance the force is applied over. Therefore, we get more work done if the object is initially moving faster.
The supposed issue behind all of this is that is seems like we are getting more energy from applying the same force for the same amount of time. But if you work through it you find that this is no issue at all. This is even true for objects falling near the Earth's surface. Even though the force is constant, gravity does more and more work on the object while it falls. Or in other words, the rate of energy conversion from potential to kinetic energy increases as the object falls.
answered 54 mins ago
Aaron Stevens
2,850319
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?
To understand this it is useful to consider a “toy model†rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).
Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning†the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.
So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?
Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.
At low speeds most of the fuel’s energy is “wasted†in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.
answered 13 mins ago
Dale
1,039211
add a comment |Â
up vote
4
down vote
The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?
To understand this it is useful to consider a “toy model†rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).
Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning†the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.
So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?
Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.
At low speeds most of the fuel’s energy is “wasted†in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.
answered 13 mins ago
Dale
1,039211
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?
To understand this it is useful to consider a “toy model†rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).
Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning†the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.
So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?
Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.
At low speeds most of the fuel’s energy is “wasted†in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.
answered 13 mins ago
Dale
1,039211
The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?
To understand this it is useful to consider a “toy model†rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).
Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning†the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.
So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?
Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.
At low speeds most of the fuel’s energy is “wasted†in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.
answered 13 mins ago
Dale
1,039211
answered 13 mins ago
Dale
1,039211
answered 13 mins ago
Dale
1,039211
answered 13 mins ago
Dale
1,039211
1,039211
add a comment |Â
add a comment |Â
up vote
2
down vote
There are multiple ways to view this.
The easiest, I think, is that kinetic energy scales with the square of velocity
$$K=frac 12 m v^2$$
If we assume that the rocket booster supplies a constant acceleration, then comparing initial and final velocities we find that
$$Delta v=v_final-v_init=at$$
So for the same amount of time, the change in the velocity is the same, regardless of what the starting velocity actually is. Since we have a squared dependence on the velocity in $K$, this means that the kinetic energy increases more if we started out with a larger velocitiy. i.e.
$$Delta K=frac 12 m v_final^2-frac 12 m v_init^2=frac 12 m (v_final-v_init)(v_final+v_init)=frac 12 m Delta v (Delta v+2v_init)$$
So as we can see, the sum of the velocities in the expression for $Delta K$ is what contributes to a larger change in kinetic energy. Since the work done is equal to the change in kinetic energy, it must be that the rocket does more work when we start off at a larger velocity.
The second way to view this, which you could argue is the same as the first, is to look at the definition of work
$$W=intvec F cdot dvec x$$
Or in one dimension with a constant force
$$W=FDelta x$$
Now, once again assuming a constant acceleration, we know that
$$Delta x = frac 12 a t^2 + v_initt$$
So that the work done is
$$W=Fleft(frac 12 a t^2 + v_inittright)$$
Once again, we see that the initial velocity determines the work. A qualitative explanation from this is that when the velocity is larger, then the object covers more distance in the same amount of time. So if we look at the time the force is being applied, the faster it is moving the larger distance the force is applied over. Therefore, we get more work done if the object is initially moving faster.
The supposed issue behind all of this is that is seems like we are getting more energy from applying the same force for the same amount of time. But if you work through it you find that this is no issue at all. This is even true for objects falling near the Earth's surface. Even though the force is constant, gravity does more and more work on the object while it falls. Or in other words, the rate of energy conversion from potential to kinetic energy increases as the object falls.
answered 54 mins ago
Aaron Stevens
2,850319
add a comment |Â
up vote
2
down vote
There are multiple ways to view this.
The easiest, I think, is that kinetic energy scales with the square of velocity
$$K=frac 12 m v^2$$
If we assume that the rocket booster supplies a constant acceleration, then comparing initial and final velocities we find that
$$Delta v=v_final-v_init=at$$
So for the same amount of time, the change in the velocity is the same, regardless of what the starting velocity actually is. Since we have a squared dependence on the velocity in $K$, this means that the kinetic energy increases more if we started out with a larger velocitiy. i.e.
$$Delta K=frac 12 m v_final^2-frac 12 m v_init^2=frac 12 m (v_final-v_init)(v_final+v_init)=frac 12 m Delta v (Delta v+2v_init)$$
So as we can see, the sum of the velocities in the expression for $Delta K$ is what contributes to a larger change in kinetic energy. Since the work done is equal to the change in kinetic energy, it must be that the rocket does more work when we start off at a larger velocity.
The second way to view this, which you could argue is the same as the first, is to look at the definition of work
$$W=intvec F cdot dvec x$$
Or in one dimension with a constant force
$$W=FDelta x$$
Now, once again assuming a constant acceleration, we know that
$$Delta x = frac 12 a t^2 + v_initt$$
So that the work done is
$$W=Fleft(frac 12 a t^2 + v_inittright)$$
Once again, we see that the initial velocity determines the work. A qualitative explanation from this is that when the velocity is larger, then the object covers more distance in the same amount of time. So if we look at the time the force is being applied, the faster it is moving the larger distance the force is applied over. Therefore, we get more work done if the object is initially moving faster.
The supposed issue behind all of this is that is seems like we are getting more energy from applying the same force for the same amount of time. But if you work through it you find that this is no issue at all. This is even true for objects falling near the Earth's surface. Even though the force is constant, gravity does more and more work on the object while it falls. Or in other words, the rate of energy conversion from potential to kinetic energy increases as the object falls.
answered 54 mins ago
Aaron Stevens
2,850319
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There are multiple ways to view this.
The easiest, I think, is that kinetic energy scales with the square of velocity
$$K=frac 12 m v^2$$
If we assume that the rocket booster supplies a constant acceleration, then comparing initial and final velocities we find that
$$Delta v=v_final-v_init=at$$
So for the same amount of time, the change in the velocity is the same, regardless of what the starting velocity actually is. Since we have a squared dependence on the velocity in $K$, this means that the kinetic energy increases more if we started out with a larger velocitiy. i.e.
$$Delta K=frac 12 m v_final^2-frac 12 m v_init^2=frac 12 m (v_final-v_init)(v_final+v_init)=frac 12 m Delta v (Delta v+2v_init)$$
So as we can see, the sum of the velocities in the expression for $Delta K$ is what contributes to a larger change in kinetic energy. Since the work done is equal to the change in kinetic energy, it must be that the rocket does more work when we start off at a larger velocity.
The second way to view this, which you could argue is the same as the first, is to look at the definition of work
$$W=intvec F cdot dvec x$$
Or in one dimension with a constant force
$$W=FDelta x$$
Now, once again assuming a constant acceleration, we know that
$$Delta x = frac 12 a t^2 + v_initt$$
So that the work done is
$$W=Fleft(frac 12 a t^2 + v_inittright)$$
Once again, we see that the initial velocity determines the work. A qualitative explanation from this is that when the velocity is larger, then the object covers more distance in the same amount of time. So if we look at the time the force is being applied, the faster it is moving the larger distance the force is applied over. Therefore, we get more work done if the object is initially moving faster.
The supposed issue behind all of this is that is seems like we are getting more energy from applying the same force for the same amount of time. But if you work through it you find that this is no issue at all. This is even true for objects falling near the Earth's surface. Even though the force is constant, gravity does more and more work on the object while it falls. Or in other words, the rate of energy conversion from potential to kinetic energy increases as the object falls.
answered 54 mins ago
Aaron Stevens
2,850319
There are multiple ways to view this.
The easiest, I think, is that kinetic energy scales with the square of velocity
$$K=frac 12 m v^2$$
If we assume that the rocket booster supplies a constant acceleration, then comparing initial and final velocities we find that
$$Delta v=v_final-v_init=at$$
So for the same amount of time, the change in the velocity is the same, regardless of what the starting velocity actually is. Since we have a squared dependence on the velocity in $K$, this means that the kinetic energy increases more if we started out with a larger velocitiy. i.e.
$$Delta K=frac 12 m v_final^2-frac 12 m v_init^2=frac 12 m (v_final-v_init)(v_final+v_init)=frac 12 m Delta v (Delta v+2v_init)$$
So as we can see, the sum of the velocities in the expression for $Delta K$ is what contributes to a larger change in kinetic energy. Since the work done is equal to the change in kinetic energy, it must be that the rocket does more work when we start off at a larger velocity.
The second way to view this, which you could argue is the same as the first, is to look at the definition of work
$$W=intvec F cdot dvec x$$
Or in one dimension with a constant force
$$W=FDelta x$$
Now, once again assuming a constant acceleration, we know that
$$Delta x = frac 12 a t^2 + v_initt$$
So that the work done is
$$W=Fleft(frac 12 a t^2 + v_inittright)$$
Once again, we see that the initial velocity determines the work. A qualitative explanation from this is that when the velocity is larger, then the object covers more distance in the same amount of time. So if we look at the time the force is being applied, the faster it is moving the larger distance the force is applied over. Therefore, we get more work done if the object is initially moving faster.
The supposed issue behind all of this is that is seems like we are getting more energy from applying the same force for the same amount of time. But if you work through it you find that this is no issue at all. This is even true for objects falling near the Earth's surface. Even though the force is constant, gravity does more and more work on the object while it falls. Or in other words, the rate of energy conversion from potential to kinetic energy increases as the object falls.
answered 54 mins ago
Aaron Stevens
2,850319
answered 54 mins ago
Aaron Stevens
2,850319
answered 54 mins ago
Aaron Stevens
2,850319
answered 54 mins ago
Aaron Stevens
2,850319
2,850319
add a comment |Â
add a comment |Â
Pilchard123 is a new contributor. Be nice, and check out our Code of Conduct.
Pilchard123 is a new contributor. Be nice, and check out our Code of Conduct.
Pilchard123 is a new contributor. Be nice, and check out our Code of Conduct.
Pilchard123 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f428952%2fwhy-is-work-done-on-an-object-greater-at-higher-speeds%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Net work done on an object such as a rocket always equals its change in kinetic energy.
– David White
1 hour ago