# Velocity

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#1

A parachutist of mass 80.0 kg drops from a plane travelling at 40.0 m/s, 2000 m above the Earth’s surface.
The parachutist hits the ground at a speed of 5.00 m s–1.
How much work is done by the parachutist against drag forces during the fall? (Take the Earth’s gravitational field strength to be 10.0 N kg–1.)

The work can be found by energy conservation, but I'm wondering why I can't get the work by kinematic equations. W = F(friction)*d
vf^2 = 0^2 + 2*(a-g)*y. That would solve for the acceleration of the friction force (a).
W = m*a*y.
Could someone please tell me what I'm doing wrong?

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1 month ago
#2
(Original post by freestyler01)

A parachutist of mass 80.0 kg drops from a plane travelling at 40.0 m/s, 2000 m above the Earth’s surface.
The parachutist hits the ground at a speed of 5.00 m s–1.
How much work is done by the parachutist against drag forces during the fall? (Take the Earth’s gravitational field strength to be 10.0 N kg–1.)

The work can be found by energy conservation, but I'm wondering why I can't get the work by kinematic equations. W = F(friction)*d
vf^2 = 0^2 + 2*(a-g)*y. That would solve for the acceleration of the friction force (a).
W = m*a*y.
Could someone please tell me what I'm doing wrong?

For your kinematic approach, you're considering vertical motion only. Whereas friction acts in the opposite direction to the direction of motion so would act horizontally to begin with (assuming the flight is level). Similarly is vf = 5 or is it resolved or ... Also, youre assuming (average) force * distance for the work done,in the vertical direction only.

So energy conservation seems like the way to go.
Last edited by mqb2766; 1 month ago
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#3
Since change in P.E = change in K.E (P.Ef - P.Ei = K.Ef - K.Ei), P.Ei - K.Ei = P.Ef - K.Ef
Defining upward direction as positive, when I do m*(-10)*h - 0.5m*vi^2 = 0.5*m*vf^2 - W, I get a slightly different answer. Why is that?
Last edited by freestyler01; 1 month ago
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1 month ago
#4
(Original post by freestyler01)
Since change in P.E = change in K.E (P.Ef - P.Ei = K.Ef - K.Ei), P.Ei - K.Ei = P.Ef - K.Ef
Defining upward direction as positive, when I do m*(-10)*h - 0.5m*vi^2 = 0.5*m*vf^2 - W, I get a slightly different answer. Why is that?
Not 100% sure of your equation, but you should get
PE + KE = KE + W
Where the left is the initial KE (40m/s), and the right is final (5m/s) and W is work done against friction.
Last edited by mqb2766; 1 month ago
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#5
(Original post by mqb2766)
Not 100% sure of your equation, but you should get
PE + KE = KE + W
Where the left is the initial KE (40m/s), and the right is final (5m/s) and W is work done against friction.
Yes, I forgot to mention I get the right answer using that. But when checking it seems that the correct way to formulate it is change in P.E = change in K.E, rather than change in P.E + change in K.E = 0. If you look at the kinematic equations we get m*a*(hf - hi) = 0.5*m*(vf^2-vi^2), which is change in P.E = change in K.E. I could be wrong, however.
Last edited by freestyler01; 1 month ago
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1 month ago
#6
(Original post by freestyler01)
Yes, I forgot to mention I get the right answer using that. But when checking it seems that the correct way to formulate it is change in P.E = change in K.E, rather than change in P.E + change in K.E = 0. If you look at the kinematic equations we get mg*(hf - hi) = 0.5*m*(vf^2-vi^2), which is change in P.E = change in K.E.
The change in PE is not the change in KE, due to the work done against friction. The total energy is conserved which includes the work done against friction. So
Work done = initial PE + initial KE - final KE
Assuming final PE is zero (if not just subtract it). There should be an example in your textbook?
Last edited by mqb2766; 1 month ago
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1 month ago
#7
(Original post by freestyler01)

A parachutist of mass 80.0 kg drops from a plane travelling at 40.0 m/s, 2000 m above the Earth’s surface.
The parachutist hits the ground at a speed of 5.00 m s–1.
How much work is done by the parachutist against drag forces during the fall? (Take the Earth’s gravitational field strength to be 10.0 N kg–1.)

The work can be found by energy conservation, but I'm wondering why I can't get the work by kinematic equations. W = F(friction)*d
vf^2 = 0^2 + 2*(a-g)*y. That would solve for the acceleration of the friction force (a).
W = m*a*y.
Could someone please tell me what I'm doing wrong?

Hi, I understand that you may have cleared your doubt.
A few questions to ask before I shared what is the issue(s) with this question and the answer provided by mqb2766.
Are you approaching this question from a physics point of view or applied mathematics (ie mechanics) point of view? Although it may seem like a redundant question as it is asked in a physics forum, based on past experiences there are students who asked about mechanics problems from A level maths in the physics forum before. I just need confirmation.
Have you learnt thermal energy?

I believe mqb2766 is answering from applied mathematics (ie mechanics) point of view.
1
#8
(Original post by mqb2766)
The change in PE is not the change in KE, due to the work done against friction. The total energy is conserved which includes the work done against friction. So
Work done = initial PE + initial KE - final KE
Assuming final PE is zero (if not just subtract it). There should be an example in your textbook?
Yeah I made a mistake there. I saw the kinematic equation m*a*(hf - hi) = 0.5*m*(vf^2-vi^2), and assumed that means change in P.E = change in K.E. change in P.E = -m*a*(hf-hi), so change in P.E + change in K.E = 0, as you mentioned.
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#9
(Original post by Eimmanuel)
Hi, I understand that you may have cleared your doubt.
A few questions to ask before I shared what is the issue(s) with this question and the answer provided by mqb2766.
Are you approaching this question from a physics point of view or applied mathematics (ie mechanics) point of view? Although it may seem like a redundant question as it is asked in a physics forum, based on past experiences there are students who asked about mechanics problems from A level maths in the physics forum before. I just need confirmation.
Have you learnt thermal energy?

I believe mqb2766 is answering from applied mathematics (ie mechanics) point of view.
Hey sorry for the late response, but I was asking from a general perspective, why the kinematic equations couldn't give me the answer. mqb2766's answer makes sense, that friction being not only vertical and dependent on the object's direction of motion, one would not be able to use a constant acceleration value.
Conservation of energy (accounting for friction/thermal energy) does give the answer, but I was simply wondering about using kinematic equation to find the work.
Last edited by freestyler01; 4 weeks ago
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1 week ago
#10
(Original post by freestyler01)
Hey sorry for the late response, but I was asking from a general perspective, why the kinematic equations couldn't give me the answer. mqb2766's answer makes sense, that friction being not only vertical and dependent on the object's direction of motion, one would not be able to use a constant acceleration value.
Conservation of energy (accounting for friction/thermal energy) does give the answer, but I was simply wondering about using kinematic equation to find the work.
First of all, I would say the following question
A parachutist of mass 80.0 kg drops from a plane travelling at 40.0 m/s, 2000 m above the Earth’s surface. The parachutist hits the ground at a speed of 5.00 m s–1. How much work is done by the parachutist against drag forces during the fall? (Take the Earth’s gravitational field strength to be 10.0 N kg–1.)

and the question in this thread
https://www.thestudentroom.co.uk/sho....php?t=7124716

are not really asking from the physics point of view.

PE + KE = KE + W
Where the left is the initial KE (40m/s), and the right is final (5m/s) and W is work done against friction.
mqb2766's answer makes sense from the perspective of mechanics viewpoint (mechanics in applied mathematics) instead of physics viewpoint.

A lot of people (including me) learn that we can calculate work done by drag force or frictional force as well as work done against drag force or frictional force. However, this work done is NOT calculable in physics.

Below is the link which describes what is wrong.
https://www.wired.com/story/even-phy...lightly-wrong/

One may argue that Rhett Allain discussed about work done by friction is not calculable NOT work done by drag force. However, both frictional force and drag force are dissipative forces that cause the macroscopic kinetic energy of a system to be “dissipated” as thermal energy. This implies that when there is work done by such forces, the system will heat up making work done by such forces NOT calculable.

Below are some physics texts that have discarded the concept of work done by frictional force or drag force:
Matter and interactions mentioned by Rhett Allain
Physics for Scientists and Engineers A Strategic Approach With Modern Physics by Randall D. Knight
Physics for Scientists and Engineers with Modern Physics by Raymond A. Serway and John W. Jewett, Jr. (from 7th edition onward)
Six Ideas that Shaped Physics by Thomas A. Moore
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