How Do You Calculate Normal Force and Speed in Physics Problems?

[cjb19]

Homework Statement

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?
Correct: Your answer is correct. m/s

(b) What is its speed at the height of 0.553 m?
Correct: Your answer is correct. m/s

(c) How much higher does it go?

Homework Equations

F=ma
v^2=v0^2+2ad

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-:/

 

[PeterO]

Homework Statement

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?
Correct: Your answer is correct. m/s

(b) What is its speed at the height of 0.553 m?
Correct: Your answer is correct. m/s

(c) How much higher does it go?

Homework Equations

F=ma
v^2=v0^2+2ad

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-:/

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

 

[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

 

[PeterO]

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

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

The problem states that K_s = 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?

 

[cjb19]

Wow, apparently I can't read.

And yes, I found that post but I think my setup is different then theirs but still correct.

I know get a value of 27.9523 N for the normal force.