Find the spring constant and speed?

In summary: E1-E2=0.5kx2+0.5kSo at the top of the spring (point A), there is potential energy (E1) that is preserved. Then at maximum compression, there is another potential energy (E2) that is also preserved because the height is less than the reference point.
  • #1
sonutulsiani
138
0

Homework Statement



A cube of mass m = 5 kg is released from rest and falls a height of h1 = 2 m before hitting a vertical spring which brings the cube to a stop (momentarily) after being compressed by h2 = 50 cm.

http://bcs.whfreeman.com/WebPub/Physics/tiplerphysics6e_bridge/question_bank_images/7-44.png?11

Assume the spring has no mass, heating of the spring and the cube is negligible, and there is no air resistance.1. Which of the following statements are true for the system which includes Earth, the cube, and the spring: (There are more than 1 true answers)

A. The mechanical energy of the system is preserved.
B. The gravitational potential energy of the cube is converted into elastic potential energy of the spring at maximum compression.
C. The kinetic energy of the cube at the time of impact is equal to the elastic potential energy of the spring at maximum compression.
D. At no point during the motion of the cube will the net force on it be zero.2. Calculate the spring constant k = ____ N/m (use g = 9.81 m/s^2 and round your answer to the nearest integer).3. What is the speed of the cube at a height h3 = 20 cm below the point of impact?

A. 6 m/s.
B. 0 m/s.
C. 21 m/s.
D. 980 m/s.

Homework Equations


The Attempt at a Solution



For the first question, I think it is A and B?
I don't know about the 2nd and 3rd question. Please help
 
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  • #2
sonutulsiani said:
For the first question, I think it is A and B?
I don't know about the 2nd and 3rd question. Please help

I believe A and B are true.

For the 2nd part, re-read what the option of B in part 1 is saying and think about what it means in terms of equations.
 
  • #3
So the equation is W = mgh and W = 0.5 kx^2 ?
So W = 5*9.81*2 = 98.1
And W = 0.5*k*(0.5m)^2 = 0.125 k ?

Is it like this
 
  • #4
How should I proceed further??
 
  • #5
At the initial height is point A.

The point at the spring in equilibrium is B (this is our reference line, so heights are positive or negative depending on where it is in relation to this point)

The point at maximum compression is C.

At what is the energy at point A?


At point C, what energies is present?
 
  • #6
At point A, there is potential energy = mgh is there.
At point C, potential energy of the spring is there = 0.5 k x^2 is present and I guess mgh is also present?
 
  • #7
sonutulsiani said:
At point A, there is potential energy = mgh is there.
At point C, potential energy of the spring is there = 0.5 k x^2 is present and I guess mgh is also present?

Yes and yes


So by conservation of energy we will have mgh1=0.5kx2+mgh2

What is h2 here knowing how we are measure the heights?
 
  • #8
H1 = 2m
H2 = 0.5 m
So then what is x? 2-0.5?
 
  • #9
sonutulsiani said:
H1 = 2m
H2 = 0.5 m
So then what is x? 2-0.5?

so mg(h1-h2)=0.5kx2


ok we know h1=2 and x=0.5.


So h1 to the bottom of the spring is 2m and the distance between B and C is 0.5m, so what is the distance of C to the bottom of the spring?
 
  • #10
How do I find that without knowing the height of the spring?
Is it 1.5? Or 0.5?
 
  • #11
If you take h1 as the height of the spring, which is 2, then h2 will be 1.5
 
  • #12
sonutulsiani said:
How do I find that without knowing the height of the spring?
Is it 1.5? Or 0.5?

sonutulsiani said:
If you take h1 as the height of the spring, which is 2, then h2 will be 1.5

Ignore my last post.


See how I said to take the reference point as point B?

at maximum compression, in relation to point B, h2=-0.5m

so your equation is mgh1=0.5kx2
mgh2 and they told you x=0.5m
 
  • #13
What? I didn't understand, what is h2 again?
How is h2= -0.5

H1 = 2, x = 0.5, and the equation is mg(h1-h2) = 0.5 k x^2 right?

I understood the H1, x and the equation part. I didn't understand h2 and how do you get it.
 
  • #14
sonutulsiani said:
What? I didn't understand, what is h2 again?
How is h2= -0.5

H1 = 2, x = 0.5, and the equation is mg(h1-h2) = 0.5 k x^2 right?

I understood the H1, x and the equation part. I didn't understand h2 and how do you get it.

ok I probably confused you with that. Let me try to clarify.

A
|
|
|
|
|
|
B (0 potential energy or reference point...h=0)
|
|
C


B is the top of the spring such that AB=2m (h1)

C is the maximum compression of the spring such that BC=0.5m


Now at A, we know E1=mgh1.

Meaning that relative to the point B, E1=mg(h1-0)

Now at maximum compression, the height relative to B is h2= -0.5m since it is on the other side of the reference point.

So the conservation of energy equation becomes mgh1=(1/2)kx2+mgh2
 
  • #15
I get it now.
So k = 981 N/m?

And for 3rd part, is it 0 m/s?
 
  • #16
sonutulsiani said:
I get it now.
So k = 981 N/m?

And for 3rd part, is it 0 m/s?

:uhh: that spring looks fishy


but anyhow, for the third part it is the same process. Now instead of the right side being elastic and graviational pe, there are these two plus what type of energy?
 
  • #17
I don't know what you said. Can you please reply soon, I have to submit in 1 hour!
 
  • #18
sonutulsiani said:
I don't know what you said. Can you please reply soon, I have to submit in 1 hour!

ok well it's like this at h=2m, there is mgh right?

this is converted into gravitation pe at the height 0.2m (relative to B),elastic potential energy of the spring and what other type of energy? (it is moving so what energy does it have?)
 
  • #19
kinetic?
 
  • #20
So how will be the equation?
 
  • #21
mgh = 1/2 k x^2 + 1/2 m v^2?
 
  • #22
Then what is h and x?
h= 2m
x = 0.2 m?
 
  • #23
If the values and the equation are right, then I get answer as 5.6 m/s.. Is it right?
 

1. How do you find the spring constant?

The spring constant, denoted by k, can be found by dividing the force applied to the spring by the displacement caused by that force. This can be represented by the equation k = F/x, where F is the force and x is the displacement.

2. What is the unit for spring constant?

The unit for spring constant depends on the unit used to measure force and displacement. If force is measured in Newtons and displacement in meters, then the unit for spring constant is N/m. If force is measured in pounds and displacement in inches, then the unit for spring constant is lbs/in.

3. How does the spring constant affect the speed of an object?

The spring constant determines the strength of the force exerted by the spring. The stronger the spring constant, the greater the force applied to the object attached to the spring. This force then affects the speed of the object, as the stronger force will result in a faster speed.

4. Can the spring constant change?

Yes, the spring constant can change depending on the material and structure of the spring. A change in the length or thickness of the spring can also affect the spring constant. Additionally, the spring constant can change if the temperature changes, as some materials may become more or less stiff at different temperatures.

5. How can you measure the speed of an object attached to a spring?

The speed of an object attached to a spring can be measured using a stopwatch and a known distance. The object's displacement should be measured when it is at rest and when it reaches its maximum displacement. The time taken for the object to complete one oscillation can then be measured, and the speed can be calculated using the equation v = 2π√(k/m), where k is the spring constant and m is the mass of the object.

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