- #1

- 318

- 0

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.

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- Thread starter Sneakatone
- Start date

- #1

- 318

- 0

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.

- #2

- 37

- 0

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

- 0

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!

Thank you!

- #5

- 37

- 0

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.

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

--- After your second comment, you are correct.

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