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How to find magnitude of initial displacement?

  • Thread starter Sneakatone
  • Start date
  • #1
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A mass of 130 g is held by a horizontal spring constant 22 N/m. It is displaced from its equilibrium position and released from the rest. As it passes through its equilibrium position, its speed is 4.2 m/s.

a) find work done on the spring.
w=1/2mv^2
1/2(.13kg)(4.2)^2= 1.14 (correct)

b) what is the magnitude of the initial displacement?
I dont know what to do for this , Im guessing were suppose to find distance but I dont know how.
 

Answers and Replies

  • #2
You have the work correct. Now you have to know how much displacement has been done for that amount of work.

The equation for the force of a spring is F = -kx, where x is the distance displaced from equilibrium. Thus, the work, which is ∫F dx = -k x^2 / 2. The negative sign signifies that the work is done on the system.

Thus, your value for work, W = k x^2 / 2, so x = √(2W/k).
 
  • #3
318
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can I do 0.13kg/22=0.005 m to get displacement?
 
  • #4
318
0
never mind x=sqrt(2*1.14/22)=0.32 m
Thank you!
 
  • #5
You are nearly right, but you forgot to include velocity. The displacement would be x = v√(m/k).

Your units do not make sense in that answer. (displacement ≠ kg / (N / m))

--- After your second comment, you are correct.
 
Last edited:

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