Spring: K.E. - P.E. relationship; Request shorter procedure

  • Thread starter wirefree
  • Start date
  • #1
102
21
Question:-
A block of mass 10 kg moving with a velocity of 1 m/s collides with a spring of force constant 1000 N/m. Calculate the compression of the spring at the moment when kinetic energy of the block is equal to one-fourth of the elastic potential energy of the spring.

Attempt:-
Step 1: Determine initial Kinetic Energy of block
Step 2: Using above value of K.E., determine maximum compression of spring using Law of Conservation of Mechanical Energy
Step 3: Determine acceleration for the maximum compression using 3rd Kinematic equation
Step 4: Obtain two equations with two unknowns (v & x):
- First: v2 = u2 + 2ax (3rd Kinematic equation)
- Second: 1/4*1/2*k*x2 = 1/2*m*v2 (from question statement)
Step 5: Substitute v2 from First equation in the Second and solve 25x2 + 10x - 1 = 0 for x.

Problem:-
The above question is a 4 marks exam question which gives me 9 minutes to solve it. In addition to the 4 steps preceding it, Step 5's quadratic equation doesn't yield whole number roots. To be precise they are x=.09 & x=-.48.

There must be a shorter procedure.

Would appreciate advise.

Regards,
wirefree
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,857
1,655
Just use conservation of energy directly. You do not need suvat equations.
The lynchpin is that the final kinetic energy is equal to one quarter of the final potential energy.
It's just one line.
 
  • #3
102
21
Appreciate the suggestion, Mr Bridge.

I fail to take lead from your suggestion.

The lynchpin is that the final kinetic energy is equal to one quarter of the final potential energy.
The final K.E. is clearly zero.

Would appreciate elaboration.

Best,
wirefree
 
  • #4
Simon Bridge
Science Advisor
Homework Helper
17,857
1,655
The final K.E. is clearly zero.
This is not what the problem statement says.
Please reread the problem statement - it tells you the final kinetic energy is not zero when it says
you said:
Calculate the compression of the spring at the moment when [the] kinetic energy of the block is equal to one-fourth of the elastic potential energy of the spring.
It does not say anywhere that the mass comes to rest.

If the final kinetic energy were zero, then the calculation would be easy - all the initial kinetic energy ends up as energy stored in the spring: ##mu^2=kx^2## ... solve for x.

If T is kinetic energy and U is potential energy, then conservation of energy says:
$$T_i+U_i=T_f+U_f$$... you know that ##U_i=0## and you have an expression for ##T_f## in terms of ##U_f## ... so substitute and simplify.
 
Last edited:
  • Like
Likes 1 person
  • #5
102
21
It does not say anywhere that the mass comes to rest.
It nevertheless would when the spring sustains maximum compression. This and this alone could be termed Final K.E. (= zero).

ImageUploadedByPhysics Forums1402120967.582200.jpg
 
  • #6
1,948
200
Yes the block stops at maximum compression. The question is not about maximum compression. It is about the compression that happens when potential energy is four times larger than kinetic energy. The problem is easier than you're making it out to be. Besides, your method will fail because you are using equations derived for motion at constant acceleration. That's not the case here. One more thing, the equation you're calling 3rd kinematic equation is actually called Torricelli's equation. There is no equation commonly known as 3rd Kinematic equation.
 
  • #7
102
21
your method will fail because you are using equations derived for motion at constant acceleration.
As shown in the graph I attach in this post, Spring Force evolves linearly. Doesn't that indicate that the Force on the spring is constant and, therefore, so is acceleration?

Interestingly, all 5 steps of my original post yield the same answer as the procedure outlined by Mr. Bridge in post #4.
 

Attachments

  • #8
1,116
72
Your OP asked for a shorter procedure.

Initial energy = 0.5 mv^2= 5J
Final energy = 5J = KE + PE =KE + 4KE
KE =1J, PE =4J= 0.5 kx^2
x = 1/5√5
 
  • #9
Simon Bridge
Science Advisor
Homework Helper
17,857
1,655
It nevertheless would when the spring sustains maximum compression. This and this alone could be termed Final K.E. (= zero).
That is correct - but nowhere does it say that the spring reaches maximum compression.

At maximum compression the final kinetic energy cannot be expressed as a fraction of the final potential energy.

As shown in the graph I attach in this post, Spring Force evolves linearly. Doesn't that indicate that the Force on the spring is constant and, therefore, so is acceleration?

Interestingly, all 5 steps of my original post yield the same answer as the procedure outlined by Mr. Bridge in post #4.
Nowhere have I said that the method you used was wrong.
You wanted a shorter approach - this is a shorter approach.
 
Last edited:
  • #10
nasu
Gold Member
3,769
428
As shown in the graph I attach in this post, Spring Force evolves linearly. Doesn't that indicate that the Force on the spring is constant and, therefore, so is acceleration?
No, it does not. It shows that the force is proportional to displacement and so is the acceleration.

Your equation (obtianed by combinig "first" and "second") should contain u and a. What values did you take for these quantities?
 
Last edited:
  • #11
Simon Bridge
Science Advisor
Homework Helper
17,857
1,655
Indeed - having taken another look:
[...] all 5 steps of my original post yield the same answer as the procedure outlined by Mr. Bridge in post #4.
Please show your working.

Considering that the assumption of constant acceleration is incorrect, it is a tad surprising.
But these things happen.
 
Last edited:
  • #12
jtbell
Mentor
15,716
3,845
As shown in the graph I attach in this post, Spring Force evolves linearly. Doesn't that indicate that the Force on the spring is constant and, therefore, so is acceleration?
The spring force varies with position.

Therefore (via F = ma) the acceleration also varies with position.

The block's position is not constant (it moves!), therefore the acceleration is not constant.

Therefore the kinematic equations for constant acceleration do not apply.
 
  • #13
228
55
Your OP asked for a shorter procedure.

Initial energy = 0.5 mv^2= 5J
Final energy = 5J = KE + PE =KE + 4KE
KE =1J, PE =4J= 0.5 kx^2
x = 1/5√5
This.

(but it is [tex]\frac{1}{5}\sqrt{2}[/tex] )
 
  • #14
1,116
72
This.

(but it is [tex]\frac{1}{5}\sqrt{2}[/tex] )
Matt, that's quite cryptic. Is that your solution for x?
:yuck:
 
  • #15
228
55
Matt, that's quite cryptic. Is that your solution for x?
:yuck:

Yes, :-)
 
  • #16
1,116
72
I've tried and tried but I cannot get your answer. Can you show me where I went wrong?
:cry:
 
  • #17
228
55
I've tried and tried but I cannot get your answer. Can you show me where I went wrong?
:cry:
[tex]4 = \frac{1}{2}k x^2[/tex] So....


...MY GOODNESS!!! I am really sorry, mate, I read k=100 but now I see it is k=1000.

Your result is correct. I was just reading too fast and saw 100 instead of 1000. My fault.

Sorry :-)
 
  • Like
Likes 1 person
  • #18
1,116
72
:smile: Thanks Matt!
 

Related Threads on Spring: K.E. - P.E. relationship; Request shorter procedure

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
2K
Replies
1
Views
2K
Replies
7
Views
820
Replies
10
Views
1K
Top