Kinematics and also an energy

In summary, the problem involves a block being sent up a frictionless ramp with a kinetic energy of 38 J. With an initial speed of 4.70 m/s, the normal force on the block can be calculated using the equations F=ma and KE=0.5mv^2. For the second question, a startled armadillo leaps upward and rises 0.553 m in the first 0.197 s. The initial speed and speed at 0.553 m can be calculated using the equation v^2=v0^2+2ad. The question also asks for the height the armadillo reaches, which can be solved by setting the initial velocity to 0 m/s at the top of the
  • #1
cjb19
3
0

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-:/
 
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  • #2
cjb19 said:

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]
 
  • #3
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
cjb19 said:
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 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
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.
 

1. What is kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion.

2. What are the three main components of kinematics?

The three main components of kinematics are position, velocity, and acceleration. Position is the location of an object in space, velocity is the rate of change of an object's position, and acceleration is the rate of change of an object's velocity.

3. How is energy related to kinematics?

Energy and kinematics are related because energy is a measure of an object's ability to do work, which is directly related to an object's motion. Kinetic energy, for example, is the energy an object possesses due to its motion.

4. What is the difference between energy and power?

Energy is the ability to do work, while power is the rate at which work is done. In other words, energy is a measure of the total amount of work done, while power is a measure of how quickly the work is being done.

5. How does the conservation of energy apply to kinematics?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed. In kinematics, this means that the total energy of a system will remain constant, even as the object's position, velocity, and acceleration change.

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