# Kinematics and also an energy

1. Dec 17, 2011

### cjb19

1. The problem statement, all variables and given/known data
1.A block is sent up a frictionless ramp along which an x axis extends upward. The figure below gives the kinetic energy of the block as a function of position x; the scale of the figure's vertical axis is set by Ks = 38.0 J. If the block's initial speed is 4.70 m/s, what is the normal force on the block?

2. A startled armadillo leaps upward rising 0.553 m in the first 0.197 s.
(a) What is its initial speed as it leaves the ground?

(b) What is its speed at the height of 0.553 m?

(c) How much higher does it go?

2. Relevant equations
F=ma

3. The attempt at a solution

For the first question

KE=.5mv^2
152 J = .5m(4.70 m/s)^2
m=13.8 kg

Height of the ramp
.5mv^2=mgh
152 J=13.8kg*9.8m/s^2*h
h=1.13 m
Solve for the angle of the ramp
sin-1(1.13m/2m) = 34 deg
N-mgcos34=0
N=13.8*9.8cos34=111 N. But this is wrong.

2nd question...
solved pt a and b, got 3.77 m/s for v0, 1.84 m/s vf

v0 is 0 m/s at the top of the height, so
0=3.77m/s ^2 +2*9.8m/s^2d
solve for d, get .725
difference is .17 m. But this is wrong.

Thank you so much! Final monday and I have a C-:/

2. Dec 17, 2011

### PeterO

Where did you get 152 Joules from? You said the energy was 38J [which only one quarter of 152]

3. Dec 17, 2011

### cjb19

Oops! Sorry, forgot to include this image
http://www.webassign.net/hrw/7-33.gif

It's a scale of 38 J so, at the top (0m) E=38J(4)=152 J

4. Dec 18, 2011

### PeterO

Where did you get the idea it is a scale 38J?

The problem states that Ks = 38J. That is not some sort of scale. They have not said 38J per division.

Also this problem was the subject of another post earlier?????

5. Dec 18, 2011