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kolbaska11 [484]
2 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]2 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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But mass of an object is directly proportional to inertia. So when the inertia is more on an object, it means that the object has more mass. For example, if there are two similar bricks, one that is made up of mortar and the other one is made of Styrofoam.

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3 years ago
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Explanation:

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