The magnitude of the force that a spring exerts (F) when it is stretched a distance x from its unstressed length is governed by Hooke's law, F = kx. (a) What are the dimensions of the force constant, k? (Use the following as necessary: M for mass, T for time, and L for length.)

Question

Braxton Lambert

Physics

19 August, 18:22

The magnitude of the force that a spring exerts (F) when it is stretched a distance x from its unstressed length is governed by Hooke's law, F = kx. (a) What are the dimensions of the force constant, k? (Use the following as necessary: M for mass, T for time, and L for length.)

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Answers (1)

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Montgomery · Answer 1

k = M / T^2

Explanation:

Hooke's law states that the lengthening of a spring is directly proportional to the force module applied to it, provided that the spring is not permanently deformed.

F = k * L

So:

k = F / L

Where:

[F] = N (Newton units)

[L] = m (meter)

[k] = N / m

N = (M * L) / T^2

[N] = (kg * m) / s^2

Finally:

[k] = (kg * m) / (m * s^2)

[k] = kg / s^2

k = M / T^2