Question

A spring with a spring constant of 0.70 N/m is stretched 1.5 m. What was the force?

Answer 1

Answer:

1.05 N

Explanation:

K = 0.7 N/m

e = 1.5 m

F = ?

from Hooke's law of elasticity

F = Ke

= 0.7×1.5

= 1.05 N

Answer 2

The force of the string is 1.05 N

The calculation can be done as follows

extension= 1.5 m

spring constant= 0.70 N/m

The force can be calculated as follows

Force= 1.5 × 0.70

= 1.05 N

Hence the force is 1.05 N

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