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A weight of 20.0 N hangs motionless from a spring, with a spring constant of 65.0 N/m. How far is the spring stretched from its original length?

0.130 m
0.308 m
0.769 m
3.25 m

Respuesta :

Answer:

0.308 m

Explanation:

Force = spring constant * change in length

F = k * Δx

F = 20 N

k = 65 N/m

Δx = ? m

20 = 65 * Δx

20/65 = Δx

Δx = 0.308 m

Cricetus

The spring stretched from its original length will be "0.308 m".

According to the question,

  • Spring constant, k = 65 N/m
  • Force, F = 20 N

The force will be:

→  [tex]Force =Spring \ constant\times Change \ in \ length[/tex]

or,

→ [tex]F = k\times \Delta x[/tex]

then,

→ [tex]\Delta x = \frac{F}{k}[/tex]

By putting the values,

        [tex]= \frac{20}{65}[/tex]

        [tex]= 0.308 \ m[/tex]

Thus the above answer is right.

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