Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it

Physics

1 answer:

Kay [80]3 years ago

3 0

Answer:

2N/cm

Step-by-step explanation:

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

$F=kx$

where, $F$ is the force which is stretching or compressing the spring,

$k$ is the spring constant; and

$x$ is the distance the spring is stretched.

Substituting the given values to find the elastic constant $k$ to get:

$F=kx$

$4=k(2)$

$k=\frac{4}{2}$

$k=2$

Therefore, the elastic constant is 2 Newton/cm.

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4 0

2 years ago

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton

pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:

$F = K * \Delta{x}$