# Easy conservation of energy problem? (long jump)

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1. Nov 22, 2015

### emmakr

1. The problem statement, all variables and given/known data
m(person) = 60kg
v1 = 3.0m/s
v2 = 0
d = 2m
y1 = -1m
y2 = 0

A person is completing a long jump. At event 1 they are 1m in the air and are moving at a rate of 3.0m/s. At event 2 they are at rest on the ground, 2m from their initial landing place. What is the stopping force required to bring them to rest? Ignore air resistance.

2. Relevant equations

Et1 = Et2
Et1 = Ek1 + Eg1
Et2 = Ek2 + Eg2 + Eint

3. The attempt at a solution
Et1 = mgy1 + 0.5mv1^2
= (60)(-9.8)(-1)+(60)(3^2)(0.5)
= 858J

Since v2 and y2 are both 0, Et2 = Eint

Eint = Fcos(theta)d
858J=Fcos180(2)
F=(858)/(-1)(2)
=-429N

What am I doing wrong?? How are Eint and Work related??? I figured that the stopping force was Ek and Eg being converted to thermal energy as friction and normal force bring the person to rest but I'm not sure...

2. Nov 23, 2015

### Bengalfan985

Wouldn't y1 = 1? Otherwise y2 = -2. But g would be positive as well, since you made both of them negative, the negatives cancelled out. Other than that, my work and answer is the same as yours.

Eint is the work done by friction to bring the runner to rest. All of the energy at Et1 is converted into kinetic energy once the person hits the ground. From there, all the energy is lost due to friction in order to bring the runner to rest. Normal force shouldn't effect it since it's acting perpendicular to the person's displacement.

3. Nov 23, 2015

### CWatters

4. Nov 23, 2015

### emmakr

The goal for us was to create a diagram to model the problem- as far as work and Eint go, doesn't work have to be in the direction of the object's motion? That's mainly where I was confused. In this case, if work is being done on the object in the -x direction, shouldn't the object have to be moving in the -x direction? It's accelerating in the -x direction but its motion is still in the +x direction. Thank you so much for your help!!

5. Nov 23, 2015

### CWatters

The work done on the person is negative...

Echange = Efinal - Einitial
= 0 - 858
= -858 joules

You can also see that from...

W = F.d

In this case force and displacement are in opposite directions so the work done on the person is negative.

Aside: In one of my diagrams he's a rubbish long jumper and jumps vertically before sinking 2m deep into the ground. Not very realistic but the maths doesn't change.

6. Nov 23, 2015

### haruspex

You are assuming that the vertical component of the landing velocity somehow got quickly converted into horizontal velocity, conserving KE.
It would be more reasonable to suppose that most of the vertical momentum got absorbed by muscles, tendons, bones and the ground very shortly after touching down. Some will get converted to forward motion, but maybe not much. But in that case there is not enough information. We would need to know the horizontal component of the 3m/s, which, as CWatters notes, could be minimal.
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