# Finding force constant, not given the value of k

• nicoleb14
In summary, the 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. The force constant of the spring is 25.50g and the block stays in contact with the spring for .15966 seconds.

#### nicoleb14

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 into f=-fx
and got f=-6.3765

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

nicoleb14 said:
... 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?

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?

Note that in using 'v$^{2}$=u$^{2}$+2as' one has to use the AVERAGE acceleration since in this case the acceleration is varying.

Last edited:
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)

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.

average acceleration is v/change in time, but i don't have the change in time? do i use the change in distance: .25m?

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.

this is what I am 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

I keep getting t = 0.42s.

nicoleb14 said:
this is what I am 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.

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).