Question

A spring has a spring constant of 65.5 N/m and it is
Stretched with a force of 15.3 N. How far will it stretch?
​

Answer 1

It would be stretched to 0.234m

According to Hooke's Law, we know that,

F = kx, where,

F is the force by which the spring is stretched,

k is the spring constant

x is the stretching distance.

Here, F = 15.3 N,

k = 65.5 N/m.

Putting the values in the expression above, we get,

x = F/k = (15.3/65.5) m

= 0.234 m

This the required answer.

Answer 2

elongation of spring would be 23 cm

given, spring constant of spring , K = 65.5 N/m

force applied on spring , F = 15.3 N

from Hooke's law, F = kx

where x is elongation of spring when force F is applied on spring of spring constant k.

now, 15.3 = 65.5 × x

⇒x = 15.3/65.5 = 0.233587786 ≈ 0.23 m = 23cm

hence elongation of spring = 23 cm

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