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helo!!! graphs, physics, kinetic energy, distance and freefall

Hi all,

I am currently doing the first year of a level physics and ive encountered a problem with the homework my teacher assigned to me:

It is about the graph in which the relationship of kinetic energy and distance during a free fall session is discussed, I just got confused with the relationship between those two.

I firstly thought that those two should have a quadratic relationship since Ek=1/2mv2, However, the answer suggested they should have a linear relationship like y=ax. I just dont know why kinetic energy will increase at a constant rate if the velocity is increased by power when the distance increases.


http://papers.xtremepapers.com/Edexcel/Advanced%20Level/Physics/1999%20Jun/Edexcel%20Physics%20Unit%201%20Answers.pdf

NO.40 is the original question. any help will be truly appreciated.
Original post by cyril1448
Hi all,

I am currently doing the first year of a level physics and ive encountered a problem with the homework my teacher assigned to me:

It is about the graph in which the relationship of kinetic energy and distance during a free fall session is discussed, I just got confused with the relationship between those two.

I firstly thought that those two should have a quadratic relationship since Ek=1/2mv2, However, the answer suggested they should have a linear relationship like y=ax. I just dont know why kinetic energy will increase at a constant rate if the velocity is increased by power when the distance increases.


http://papers.xtremepapers.com/Edexcel/Advanced%20Level/Physics/1999%20Jun/Edexcel%20Physics%20Unit%201%20Answers.pdf

NO.40 is the original question. any help will be truly appreciated.


Question 1

Firstly, let's consider how the velocity varies with distance. Using the relationship v2=u2+2asv^2=u^2+2as, we find that v2∝sv^2 \propto s since u = 0 and 2a is a constant. We also know that KE=12mv2K_E=\frac{1}{2}mv^2. Since v2∝sv^2 \propto s, we can substitute this into the kinetic energy equation to get KE∝sK_E \propto s. That's the mathematical explanation for why they are directly proportional (and hence the straight line graph).

The physical explanation is that (almost) all of the gravitational potential energy lost as the body falls is converted into kinetic energy. Since GPE∝sGP_E \propto s, we can say that KE∝sK_E \propto s.

Question 2

We have the relationship s=ut+12at2s=ut+\frac{1}{2}at^2. As u = 0 and 0.5a is constant, we have s∝t2s \propto t^2. Therefore, the graph is quadratic.

Question 3

Whilst the gravitational field strength isn't technically constant (it will actually increase slightly as the body drops), over small distances this is pretty much negligible so we can assume the acceleration is constant. Therefore, we'll have a straight line of gradient 0 (since we're modelling the field strength as being constant). The equation of this line will be approximately y=9.81y=9.81 but all we need to know for this question is that y > 0.

Does that make sense to you?
(edited 9 years ago)
Reply 2
Avatar for SkyWar
SkyWar
Mate wtf is this, my head hurts.


Posted from TSR Mobile
Original post by cyril1448
Hi all,

I am currently doing the first year of a level physics and ive encountered a problem with the homework my teacher assigned to me:

It is about the graph in which the relationship of kinetic energy and distance during a free fall session is discussed, I just got confused with the relationship between those two.

I firstly thought that those two should have a quadratic relationship since Ek=1/2mv2, However, the answer suggested they should have a linear relationship like y=ax. I just dont know why kinetic energy will increase at a constant rate if the velocity is increased by power when the distance increases.


http://papers.xtremepapers.com/Edexcel/Advanced%20Level/Physics/1999%20Jun/Edexcel%20Physics%20Unit%201%20Answers.pdf

NO.40 is the original question. any help will be truly appreciated.



The reason is given in the answer scheme.
As the body falls its kinetic energy increases. Its potential energy decreases.
The potential energy is converted to kinetic energy.
The potential energy = mgh and so just depends on h, the distance (m and g are constant)
So for a graph of energy (potential or kinetic) against distance for a falling object, the value is directly proportional to h and is therefore a straight line. One increasing linearly (ke) and the other decreasing linearly (pe), the total remaining constant in the absence of friction.
(edited 9 years ago)
Reply 4
Avatar for cyril1448
cyril1448
OP
Original post by Stonebridge
The reason is given in the answer scheme.
As the body falls its kinetic energy increases. Its potential energy decreases.
The potential energy is converted to kinetic energy.
The potential energy = mgh and so just depends on h, the distance (m and g are constant)
So for a graph of energy (potential or kinetic) against distance for a falling object, the value is directly proportional to h and is therefore a straight line. One increasing linearly (ke) and the other decreasing linearly (pe), the total remaining constant in the absence of friction.


Get it thank you so much

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