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myrzilka [38]
3 years ago
8

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

ion while momentarily coming to rest. If the initial speed of the box were doubled, how far x2 would the spring compress?
Physics
1 answer:
4vir4ik [10]3 years ago
3 0

Answer:

<em>2</em>x_{1}<em>, The spring would compress twice the length of </em>x_{1}<em>. </em>

Explanation:

The Spring force is an application of Hooke's law,  and using the law of conservation of energy in a spring system, we can find the relationship between the compression length and velocity.

<h3>Step by Step Calculations</h3>

For a spring system, the conservation of energy can be applied using the formula below;

\frac{1}{2} kx_{1} ^{2}  = \frac{1}{2} mv^{2} ........................................1

Where k is the spring constant

x_{1} is the distance

m is the mass and

v is the velocity

We have to make the distance the subject formula to know how it affects the velocity;

x_{1} ^{2}  = (\frac{m}{k} )v^{2}

x_{1} = \sqrt{(\frac{m}{k} )}v .....................................2

from the equation, it shows that the distance (x_{1}) is directly proportional to the velocity (v). This implies that when the distance increases the velocity also increases.

x ∝ v

Therefore when the initial speed is doubled to 2 v it will amount a double increase in the distance which is 2x_{1}

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