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EleoNora [17]
4 years ago
11

A spring with a mass on the end of it hangs in equilibrium a distance of 0.4200 m above the floor. The mass is pulled down a dis

tance 0.0600 m below the original position, released, and allowed to oscillate. How high above the floor is the mass at the highest point in its oscillation
Physics
1 answer:
shtirl [24]4 years ago
5 0

Answer:

0.48 m

Explanation:

I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others

If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.

0.42 m + 0.06 m = 0.48 m

At the highest point thus, the height is 0.48 m above the ground.

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Hi there!

We know that:

U (Potential energy) = mgh

We are given the potential energy, so we can rearrange to solve for h (height):

U/mg = h

g = 9.81 m/s²

m = 30 g ⇒ 0.03 kg

0.062/(0.03 · 9.81) = 0.211 m

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4 years ago
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