If a 0.1 kg mass is suspended at rest from a spring near the Earth\'s surface, t

Question

If a 0.1 kg mass is suspended at rest from a spring near the Earth's surface, the distance that the spring is stretched is measured to be 1.0 cm. What is the spring constant of the spring (remember the MKS units)? A mass of 2 kg is attached to a spring with constant 18 N/m. How far is the spring stretched? A spring with a spring constant of 28.9 N/m is attached to different masses, and is stretched 5cm. What is the force exerted on the spring? What does the term "elastic limit" mean? What happens when you break this limit? What does "k" represent for a spring? Why is there a negative sign in the Hooke's Law formula?

Explanation / Answer

1) m= 0.1 kg x = 1.0 cm = 0.01 m

F =-kx

mg = - k x

k = 0.1 * 9.8 / 0.01 = 98 N/m

2) 2 * 9.8 = 18 * X

x = 1.088 m

(3) k= 28.9 x = 5cm = 0.05 m

F = -28.9 * 0.05 = 1.445

(4) Elastic limit refers the amount of stretching where stretch is proportional to stress applied.

When it breaks, after the removal it cannot restore itself

(5) K refers to the proportionality constant

(6) because displacement and force are in opposite direction

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