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A 1.6kg mass hangs at rest from a spring with a spring constant of 34N/m. How far is the spring stretched?

User Sodd
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1 Answer

4 votes

Answer:

0.46 m

Step-by-step explanation:

By Hooke's law, the applied force on the spring is equal to the spring constant times the change in length.

F = kx

Since a 1.6 kg mass hangs from the spring, the force applied is the weight of this mass, so

F = mg

F = (1.6 kg)(9.8 m/s²)

F = 15.68 N

Then, we can solve the initial equation for x and replace the values

x = F/k

x = 15.68/34

x =0.46 m

Therefore, the spring is stretched 0.46 m

User Ktutnik
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