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Falling - is the gain in KE linear?
Clash Royale CLAN TAG#URR8PPP
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When a ball falls from a high place, it experiences a gravitational force. Forces make objects accelerate (in this case, it is constantly increasing the velocity). Because $KE = frac12mv^2$ this should mean that Kinetic Energy should grow quadratically (please correct me if wrong) because of the increasing velocity right?
But also, $GPE=mgh$ where the potential energy is a linear equation. How can this happen? If energy has to be conserved wouldn't both equations have to change linearly?
Can you please explain how the $KE$, $GPE$ and conservation of energy are reconciled in this system? Could you also confirm the shape of the graph of $KE$ and $GPE$ against time?
(I had initially come up with this problem for electric fields but I think that it might've been easier to answer the question in terms of gravitational fields)
energy gravity energy-conservation potential
asked 2 hours ago
John Hon
23329
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up vote
3
down vote
favorite
When a ball falls from a high place, it experiences a gravitational force. Forces make objects accelerate (in this case, it is constantly increasing the velocity). Because $KE = frac12mv^2$ this should mean that Kinetic Energy should grow quadratically (please correct me if wrong) because of the increasing velocity right?
But also, $GPE=mgh$ where the potential energy is a linear equation. How can this happen? If energy has to be conserved wouldn't both equations have to change linearly?
Can you please explain how the $KE$, $GPE$ and conservation of energy are reconciled in this system? Could you also confirm the shape of the graph of $KE$ and $GPE$ against time?
(I had initially come up with this problem for electric fields but I think that it might've been easier to answer the question in terms of gravitational fields)
energy gravity energy-conservation potential
asked 2 hours ago
John Hon
23329
1
Linear with respect to what? Linear with respect to time - no. With respect to velocity - no. With respect to the distance fallen - yes!
– alephzero
2 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
When a ball falls from a high place, it experiences a gravitational force. Forces make objects accelerate (in this case, it is constantly increasing the velocity). Because $KE = frac12mv^2$ this should mean that Kinetic Energy should grow quadratically (please correct me if wrong) because of the increasing velocity right?
But also, $GPE=mgh$ where the potential energy is a linear equation. How can this happen? If energy has to be conserved wouldn't both equations have to change linearly?
Can you please explain how the $KE$, $GPE$ and conservation of energy are reconciled in this system? Could you also confirm the shape of the graph of $KE$ and $GPE$ against time?
(I had initially come up with this problem for electric fields but I think that it might've been easier to answer the question in terms of gravitational fields)
energy gravity energy-conservation potential
asked 2 hours ago
John Hon
23329
When a ball falls from a high place, it experiences a gravitational force. Forces make objects accelerate (in this case, it is constantly increasing the velocity). Because $KE = frac12mv^2$ this should mean that Kinetic Energy should grow quadratically (please correct me if wrong) because of the increasing velocity right?
But also, $GPE=mgh$ where the potential energy is a linear equation. How can this happen? If energy has to be conserved wouldn't both equations have to change linearly?
Can you please explain how the $KE$, $GPE$ and conservation of energy are reconciled in this system? Could you also confirm the shape of the graph of $KE$ and $GPE$ against time?
(I had initially come up with this problem for electric fields but I think that it might've been easier to answer the question in terms of gravitational fields)
energy gravity energy-conservation potential
energy gravity energy-conservation potential
asked 2 hours ago
John Hon
23329
asked 2 hours ago
John Hon
23329
asked 2 hours ago
John Hon
23329
asked 2 hours ago
John Hon
23329
asked 2 hours ago
John Hon
23329
23329
1
Linear with respect to what? Linear with respect to time - no. With respect to velocity - no. With respect to the distance fallen - yes!
– alephzero
2 hours ago
add a comment |Â
1
Linear with respect to what? Linear with respect to time - no. With respect to velocity - no. With respect to the distance fallen - yes!
– alephzero
2 hours ago
1
1
Linear with respect to what? Linear with respect to time - no. With respect to velocity - no. With respect to the distance fallen - yes!
– alephzero
2 hours ago
Linear with respect to what? Linear with respect to time - no. With respect to velocity - no. With respect to the distance fallen - yes!
– alephzero
2 hours ago
add a comment |Â
2 Answers
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You are mixing up two different SUVAT equations. The change of velocity with time is given by:
$$ v = u + at $$
So velocity increases linearly with time. However the change of velocity with distance is given by:
$$ v^2 = u^2 + 2as $$
So velocity increases as the square root of distance, not linearly with distance. That's why the kinetic energy increases linearly with distance. The kinetic energy does increase quadratically with time.
answered 2 hours ago
John Rennie
265k41517765
but then if KE increases quadratically with time, what shape must the GPE decrease with to ensure conservation of energy?
– John Hon
2 hours ago
@JohnHon GPE decreases (becomes more negative) linearly with distance and quadratically with time. This ensures that the sum of the GPE and KE remains constant.
– John Rennie
1 hour ago
add a comment |Â
up vote
0
down vote
GPE will not decrease linearly over time. This is because height will decrease exponentially as velocity increases due to acceleration (until either h is zero or terminal velocity is reached).
Otherwise, if in doubt; chart it and see.
answered 1 hour ago
user43685
434
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You are mixing up two different SUVAT equations. The change of velocity with time is given by:
$$ v = u + at $$
So velocity increases linearly with time. However the change of velocity with distance is given by:
$$ v^2 = u^2 + 2as $$
So velocity increases as the square root of distance, not linearly with distance. That's why the kinetic energy increases linearly with distance. The kinetic energy does increase quadratically with time.
answered 2 hours ago
John Rennie
265k41517765
but then if KE increases quadratically with time, what shape must the GPE decrease with to ensure conservation of energy?
– John Hon
2 hours ago
@JohnHon GPE decreases (becomes more negative) linearly with distance and quadratically with time. This ensures that the sum of the GPE and KE remains constant.
– John Rennie
1 hour ago
add a comment |Â
up vote
2
down vote
You are mixing up two different SUVAT equations. The change of velocity with time is given by:
$$ v = u + at $$
So velocity increases linearly with time. However the change of velocity with distance is given by:
$$ v^2 = u^2 + 2as $$
So velocity increases as the square root of distance, not linearly with distance. That's why the kinetic energy increases linearly with distance. The kinetic energy does increase quadratically with time.
answered 2 hours ago
John Rennie
265k41517765
but then if KE increases quadratically with time, what shape must the GPE decrease with to ensure conservation of energy?
– John Hon
2 hours ago
@JohnHon GPE decreases (becomes more negative) linearly with distance and quadratically with time. This ensures that the sum of the GPE and KE remains constant.
– John Rennie
1 hour ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You are mixing up two different SUVAT equations. The change of velocity with time is given by:
$$ v = u + at $$
So velocity increases linearly with time. However the change of velocity with distance is given by:
$$ v^2 = u^2 + 2as $$
So velocity increases as the square root of distance, not linearly with distance. That's why the kinetic energy increases linearly with distance. The kinetic energy does increase quadratically with time.
answered 2 hours ago
John Rennie
265k41517765
You are mixing up two different SUVAT equations. The change of velocity with time is given by:
$$ v = u + at $$
So velocity increases linearly with time. However the change of velocity with distance is given by:
$$ v^2 = u^2 + 2as $$
So velocity increases as the square root of distance, not linearly with distance. That's why the kinetic energy increases linearly with distance. The kinetic energy does increase quadratically with time.
answered 2 hours ago
John Rennie
265k41517765
answered 2 hours ago
John Rennie
265k41517765
answered 2 hours ago
John Rennie
265k41517765
answered 2 hours ago
John Rennie
265k41517765
265k41517765
but then if KE increases quadratically with time, what shape must the GPE decrease with to ensure conservation of energy?
– John Hon
2 hours ago
@JohnHon GPE decreases (becomes more negative) linearly with distance and quadratically with time. This ensures that the sum of the GPE and KE remains constant.
– John Rennie
1 hour ago
add a comment |Â
but then if KE increases quadratically with time, what shape must the GPE decrease with to ensure conservation of energy?
– John Hon
2 hours ago
@JohnHon GPE decreases (becomes more negative) linearly with distance and quadratically with time. This ensures that the sum of the GPE and KE remains constant.
– John Rennie
1 hour ago
but then if KE increases quadratically with time, what shape must the GPE decrease with to ensure conservation of energy?
– John Hon
2 hours ago
but then if KE increases quadratically with time, what shape must the GPE decrease with to ensure conservation of energy?
– John Hon
2 hours ago
@JohnHon GPE decreases (becomes more negative) linearly with distance and quadratically with time. This ensures that the sum of the GPE and KE remains constant.
– John Rennie
1 hour ago
@JohnHon GPE decreases (becomes more negative) linearly with distance and quadratically with time. This ensures that the sum of the GPE and KE remains constant.
– John Rennie
1 hour ago
add a comment |Â
up vote
0
down vote
GPE will not decrease linearly over time. This is because height will decrease exponentially as velocity increases due to acceleration (until either h is zero or terminal velocity is reached).
Otherwise, if in doubt; chart it and see.
answered 1 hour ago
user43685
434
add a comment |Â
up vote
0
down vote
GPE will not decrease linearly over time. This is because height will decrease exponentially as velocity increases due to acceleration (until either h is zero or terminal velocity is reached).
Otherwise, if in doubt; chart it and see.
answered 1 hour ago
user43685
434
add a comment |Â
up vote
0
down vote
up vote
0
down vote
GPE will not decrease linearly over time. This is because height will decrease exponentially as velocity increases due to acceleration (until either h is zero or terminal velocity is reached).
Otherwise, if in doubt; chart it and see.
answered 1 hour ago
user43685
434
GPE will not decrease linearly over time. This is because height will decrease exponentially as velocity increases due to acceleration (until either h is zero or terminal velocity is reached).
Otherwise, if in doubt; chart it and see.
answered 1 hour ago
user43685
434
answered 1 hour ago
user43685
434
answered 1 hour ago
user43685
434
answered 1 hour ago
user43685
434
434
add a comment |Â
add a comment |Â
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1
Linear with respect to what? Linear with respect to time - no. With respect to velocity - no. With respect to the distance fallen - yes!
– alephzero
2 hours ago