Question

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed position while momentarily coming to rest. If the initial speed of the box were doubled, how far x2 would the spring compress?

Answer 1

Answer:

2$x_{1}$, The spring would compress twice the length of $x_{1}$.

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.

Step by Step Calculations

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 2$x_{1}$