# Motion along an inclined plane

1. Feb 5, 2006

### nick727kcin

how can i possibly find this out without the mass of the boy?

Johnny jumps off a swing, lands sitting down on a grassy 20 degree slope, and slides 3.5m 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 initial speed on the grass?

i set the problem up:

i know i have to start by finding the acceleration, which i can get by finding out the force opposite of the frictional force (i forgot to include it)

F(II)= mgsin(20)

problem is, i dont know the mass of the boy

Last edited: Feb 5, 2006
2. Feb 5, 2006

### Hootenanny

Staff Emeritus
Try thinking about kinematic equations,find acceleration. Then think how this relates to force.

3. Feb 5, 2006

### Hootenanny

Staff Emeritus
HINT: Use the right equations and the masses will cancel.

4. Feb 5, 2006

### nick727kcin

thanks for trying but i still dont get it

5. Feb 5, 2006

### nick727kcin

and about the kinematic equations, it wont work because i only have distance and vf

6. Feb 5, 2006

### Hootenanny

Staff Emeritus
Did you not have any other information, such as the height of the swing etc?

7. Feb 5, 2006

### nick727kcin

none whatsoever, just what i posted

8. Feb 5, 2006

### nick727kcin

uk= .5
vi=?
vf=0 m/s
g=9.8 m/s^2
x= 3.5m
theta= 20 degrees

9. Feb 5, 2006

### Hootenanny

Staff Emeritus
I've figured it out. I'll start for you
$$F = \mu R$$
$$ma = \mu mg \cos 20$$

You take it from there.

10. Feb 5, 2006

### nick727kcin

what does that R stand for (Fn??)

and thanks alot man

11. Feb 5, 2006

### Staff: Mentor

If one uses the conservation of energy, there are three things to consider, the initial kinetic energy, the gravitational potential energy, and the energy dissipated by friction. The mass is the same in all cases, so it divides out of the energy equation, and one is left with specfic energies, i.e. independent of mass. For friction, the energy dissipated is simply force * distance and force is related to $\mu$mg*cos$\theta$ where $\theta$ is the angle between vertical and the normal to the slope.

12. Feb 5, 2006

### Hootenanny

Staff Emeritus
Yes R is the normal reaction force.

13. Feb 5, 2006

### nick727kcin

o i see now, thanks soooo much guys

i still dont get why my teaacher gave this to us thoug

14. Feb 5, 2006

### nick727kcin

ok so heres what i got

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

a=.5(9.8)cos20
a=4.604

so, i then used a kinematic equation

0=vi^2 + (-32.23)
vi= 5.677

but this is wrong......

15. Feb 5, 2006

### Hootenanny

Staff Emeritus

16. Feb 5, 2006

### nick727kcin

i dont know... the hw only tells me when i get a wrong answer (and deducts points for each try)

17. Feb 5, 2006

### Hootenanny

Staff Emeritus
This is how I would have done it, but you might want to try astronuc's method.

18. Feb 5, 2006

### nick727kcin

ok, thanks

19. Feb 5, 2006

### nick727kcin

astronuc, can you please simplify this? i havent really learned the conservation of energy law yet