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raketka [301]
3 years ago
5

A force of 34N stretches a very light ideal spring 0.73 m from equilibrium, What is the force constant (spring constant) of the

spring? (A) 47N/m (B) 38N/m (C) 53N/m (D) 25N/m
Physics
1 answer:
Lesechka [4]3 years ago
3 0

Answer:

Spring constant, k = 47 N/m

Explanation:

It is given that,

Force applied to a spring, F = 34 N

A very light ideal spring moves 0.73 m from equilibrium position i.e. x = 0.73 m

We have to find the force constant or spring constant of the spring. It can be calculated using Hooke's law. According to him, the force acting on the spring when it compresses or stretches is given by :

F=-kx  (-ve sign shows opposite direction)

k=\dfrac{F}{x}

k=\dfrac{34\ N}{0.73\ m}

k = 46.5 N/m

or

k = 47 N/m

Hence, the spring constant of the spring is 47 N/m.

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Situation B

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