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A spring with spring constant 58N/cm is stretched 4cm. How much force is it applying

User Muhamad
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1 Answer

6 votes

Answer:

232 N

Step-by-step explanation:

By Hooke's law, the force applied to a spring is proportional to the stretch of the spring, so

F = kx

Where F is the force, k is the spring constant and x is how much it is stretched.

So, replacing k by 58N/cm and x by 4 cm, we get

F = (58 N/cm)(4 cm)

F = 232 N

Therefore, the force applied is 232 N

User Roman Susi
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6.4k points