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kolbaska11 [484]
3 years ago
13

A 5.0 kg mass is suspended from a spring. Pulling the mass down by an additional 10 cm takes a force of 20 N. If the mass is the

n released, it will rise up and then come back down. How long will it take for the mass to return to its starting point 10 cm below its equilibrium position?
Physics
1 answer:
Rudik [331]3 years ago
4 0

Answer:

0.99 seconds

Explanation:

The problem depicts a simple harmonic motion.

Now, from Hooke's law,

The spring constant is given as;

k =F/x

where:

x is the displacement of the spring's end from its equilibrium position

F is the restoring force exerted by the spring on that end

From the question, F = 20N while x = 10cm = 0.1m. Thus,

K = 20/0.1 = 200 N/m

Now, time to take for the mass to return to its starting point is a period

The period oscillation of the mass is given as;

T = 2π√m/k

Where m = mass = 5kg

T = 2π√(5/200)

T = 0.99 s

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