3 years ago

When you stretch a spring 20 cm past its natural length, it exerts a force of 8

Physics

2 answers:

Nastasia [14]3 years ago

6 0

Answer:

40 N/m

Explanation:

F = -kx (This is the Hooke's Law equation)

F is the force the spring exerts = 8 N

-k = spring constant

x = displacement (The distance stretched past it's natural length) = 20cm

x needs to be in meters, and 20 cm is = to 0.2 meters

Finally:

8N = -k (0.2m)

-k = 8N / 0.2 m

k = -40 N/m

Lina20 [59]3 years ago

5 0

Answer:

0.4 N/cm

Explanation:

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