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Finding force constant, not given the value of k

  • Thread starter nicoleb14
  • Start date
  • #1
11
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A 0.650 kg block slides on a frictionless, horizontal surface with a speed of 1.19 m/s. The block encounters an unstretched spring and compresses it 25 cm before coming to rest.
(a) What is the force constant of this spring? ____ N/m

(b) For what length of time is the block in contact with the spring before it comes to rest? ___s

For part a:
Since the formula for a force constant is f=-kx, and the problem doesn't give a k value, I tried solving for it using:

k=(mg)/x and got
k=(.650kg*9.81)/.25 m
k=25.50g

and plugged this in to f=-fx
and got f=-6.3765

For part b:
im not even sure where to start ..
 

Answers and Replies

  • #2
993
13
... k=(mg)/x ...
I do not think that is correct. The force compressing the spring is NOT the weight of the mass mg.

Try thinking about conservation of energy.
What is the formula for elastic PE in a spring?
 
  • #3
11
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I got the answer for part a:

i used
v^2=u^2+2as
and just continued on from there..

how do i start doing part b?
 
  • #4
993
13
Note that in using 'v[itex]^{2}[/itex]=u[itex]^{2}[/itex]+2as' one has to use the AVERAGE acceleration since in this case the acceleration is varying.
 
Last edited:
  • #5
11
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I used v^2=u^2as
then resultant force = k X extension
.650*(-2.8322)-0=k(.25)
k=14.7

(i was originally reviewing this problem and also given this answer, but I was having a hard time getting to the answer/didn't know how to start, the answer given for part a was 14.7)
 
  • #6
993
13
For (b) one can use 'v = u + at' . But be careful to use the AVERAGE acceleration since
'v = u + at' can only be used when acceleration is constant or when the average acc is being used.
 
  • #7
11
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average acceleration is v/change in time, but i dont have the change in time? do i use the change in distance: .25m?
 
  • #8
993
13
You know v and you also know a=2.8322m/(s^2) and so you can find t. But use the average acceleration which in our case is a/2 since the force is directly proportional to the extension.
 
  • #9
11
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this is what im doing:

v=u + at
1.4161 = 1.19 + 1.4161t
.2261=1.4161t
t=.15966

but the correct answer is said to be t=.33 s
 
  • #10
993
13
I keep getting t = 0.42s.
 
  • #11
993
13
this is what im doing:

v=u + at
1.4161 = 1.19 + 1.4161t
.2261=1.4161t
t=.15966

but the correct answer is said to be t=.33 s
note that v = 0 and average acc is 2.8269m/s^2.
 
  • #12
993
13
Now I notice what I did wrong.
The acceleration is not constant since the displacement is one quarter of an shm. So the time is one quarter of the period and the period is 2piSquareRoot(m/k).
 

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