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Reptile [31]
3 years ago
10

A spring with a spring constant value of 125 N/m is compressed 12.2 cm by pushing on it with a 215 g block. When the block is re

leased, what velocity will the block have when it leaves the spring (we're ignoring friction here)? a​
Physics
1 answer:
allsm [11]3 years ago
3 0

Answer:

v = 2.94 m/s

Explanation:

When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.

Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.

Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means

(1/2)kx^2 = (1/2)mv^2

kx^2 = mv^2

v^2 = (kx^2)/m

v = sqrt((kx^2)/m)

v = x * sqrt(k/m)

v = 0.122 * sqrt(125/0.215)        <--- units converted to m and kg

v = 2.94 m/s

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a) The velocity is 2.94m/s

b) 0.441

Explanation:

a) Assume gravity is 9.8m/s^2

Use the equation below to solve for the velocity at 0.30 seconds

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vf=0m/s+(9.8m/s^2)*0.30seconds

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vf=2.94m/s

b)

Use the equation below to solve for the displacement at 0.30 seconds

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Step 1: Substitute the same variables with the knowns

x=\frac{1}{2}*(9.8m/s^2)*(0.30seconds)^{2}

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Step 2: Solve

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