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Ostrovityanka [42]
3 years ago
6

A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c

m. What is the value of the spring constant of this spring?
Physics
1 answer:
mote1985 [20]3 years ago
7 0

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.

F ∝ e

F = ke

where;

k is the spring constant

k = F / e

k = 40 / 0.04

k = 1000 N/m

Therefore, the value of the spring constant of this spring is 1000 N/m

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a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

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unknown:

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x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}

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