Determining Compression of a Spring involving Spring Constants

AI Thread Summary
To determine the compression of a spring when a 10g marble is launched at 40 cm/s, the relationship between kinetic energy and spring potential energy is used. The equations F = kx and E = 1/2 kx^2 are essential for solving the problem. The kinetic energy of the marble is calculated using the formula 1/2 mv^2, leading to the equation 0.01(0.4)^2 = 20x^2. The resulting calculation shows that the spring is compressed by 8.9 cm. The initial velocity should be corrected to 40 cm/s for accurate results.
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Homework Statement


A 10g marble is launched horizontally from a spring of K = 20 N/m. If the marble ended up with a launch speed of 40m/sec, how much (in cm) was the spring compressed by?

Homework Equations



F=kx
E=\frac{1}{2}kx^2

The Attempt at a Solution


To find how much the spring was compressed by, I know that you must solve for x in either equation. But I am unsure what values F or E would be. F =ma, but no acceleration, displacement, or time interval is given. ΔE= FΔd, but still force is required.
 
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The elastic potential energy of the spring when compressed to a distance 'x' will be converted into kinetic energy.

Do you know how to find the kinetic energy of an object given its mass and velocity?
 
\frac{1}{2}mv^2 = \frac{1}{2}kx^2

(0.01)(0.4)^2 = 20x^2

0.0089 m = x

8.9 cm = x

I wrote that the initial velocity was 40m/s, should have been 40cm/s. Would this solution be correct though?
 
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