# Johnny Jumps Off The Swing

1. Oct 27, 2007

### TonkaQD4

Johnny jumps off a swing, lands sitting down on a grassy 20 degree slope, and slides 3.5 meters down the slope before stopping. The coefficient of kinetic friction between grass and the seat of Johnny's pants is 0.5. What was his intial speed on the grass?

I started to do this problem on my own, and came to a stop due to confusion, so I looked for other forums and found a posting that was similar to the way I started my problem and then followed his finishing steps which I thought were accurate but, turns out, it is wrong and I can't figure out why. What am I doing wrong?

ma = \mu mg \cos 20

a=\mu g \cos20 (since the m's cancel out)

a=.5(9.8)cos20

a=4.604

Kinematic Equation

vf^2= vi^2 + 2ad ---> Velocity and Displacement

0=vi^2 + (-32.23)

vi= 5.677

This is wrong though...

2. Oct 27, 2007

### Oerg

In your first equation, shouldnt it be ma=mgcos20-umgcos20 instead? where u is the coefficient of friction.

3. Oct 27, 2007

### PhanthomJay

You forgot to include the component of johnny's weight down the slope in your calc for acceleration.

4. Oct 27, 2007

### TonkaQD4

We are not given Johnny's weight

5. Oct 27, 2007

### PhanthomJay

Yes, this is true. But does it matter? You have already noted that the m's cancel.....

6. Oct 28, 2007

### FedEx

The usual acc of a block on a plane of inclination theta which is at rest is given by

$$mgsin\theta\ - \mu}mgcos\theta\ = ma$$

But here it has some initial velocity and hence its acc would be different. Hence we cannot apply the above equation.

So in the above equation on the right hand side we will add mA. Where A is the acc by which the boy lands on the plane.

This mA is the resolution of mg and mA' where A' is the acc in the horizontal direction .But we are not given the type of jump which the boy commits.

Hence i think that some data is missing.

If the boy is just falling on the ground then we can easily find the answer.

7. Oct 28, 2007

### TonkaQD4

No other INFO.

I think it is just assumed that the boy falls exactly parallel to the 20 degree slope and instantly starts sliding.

There is no info that states the boy's mass.

8. Oct 28, 2007

### TonkaQD4

FED EX:

You equation actually worked if you just cancel out all the m's and then plug and chug.

g sintheta - mu g costheta = a

9.8sin20 - 0.5(9.8)cos20 = a

3.35 - 4.604 = a

a = - 1.25

Now plug this into the Kinematic equation

0 = V_i^2 + 2(-1.25)(3.5)

V_i^2 = sqrt 8.77

Initial Velocity = 2.96m/s

9. Oct 28, 2007

### TonkaQD4

How do you draw a Free Body Diagram of this situation?

10. Oct 28, 2007

### FedEx

I know we can get the answer from the equation which i gave. But the answer is wrong.Because this is for a mass which is stationary in the initial stage.

When the boy falls he has some acc and we have to consider it.This acc adds up with mgsin(theta)

11. Oct 30, 2007

### TonkaQD4

Can somebody draw a Free-Body Diagram of this situation?

12. Oct 31, 2007

### TonkaQD4

It seems like something is off for my x-componenet.

F_x=ma_x

-f_k - Fgcostheta = ma_x
-f_k - mgcostheta = ma_x
-u_kn - mgcostheta = ma_x Substitute mgsintheta in for "n" which is the y comp.
-u_k * mgsintheta - mgcostheta = ma_x u is really mu but I am writing it as "u" to simplify

Therefore

a_x = -g(u_ksintheta + costheta) ---> I'm not sure if my neg/pos are correct

-9.8(.5sin70 + cos70)

13. Nov 1, 2007

### white 2.5rs

Tonka, you go to CWU huh :)?

14. Nov 1, 2007

### PhanthomJay

I see you've switched your thetas from 20 degrees wrt x axis to 70 degrees wrt y axis, which is OK, but it confuses the problem. Then your plus/minus sign seem messed. Stick with what was posted earlier.....mgsintheta - umgcostheta =ma, where theta is 20 degrees, and a comes out negative implying the acceleration is up the plane , slowing johnny to a stop. The acceleration is independent of what acceleration johnny may have had as he smoothly hit the slopes.