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Fantom [35]
3 years ago
6

A block of mass m is initially moving to the right on a horizontal frictionless surface at a

Physics
2 answers:
slava [35]3 years ago
7 0

Answer:

The distance that the spring compresses is:

v\sqrt{\frac{m}{k}}

Explanation:

<u>Kinetic and Elastic Potential Energy</u>

The kinetic energy of an object of mass m traveling at a speed v is:

\displaystyle K=\frac{1}{2}mv^2

The elastic potential energy of a spring of constant k that compresses a distance x is:

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

The block of mass m is moving at a speed v when compresses a spring of constant k. The kinetic energy will eventually transform into elastic energy, but before that, both energies will be equal. It happens when:

\displaystyle \frac{1}{2}mv^2=\frac{1}{2}kx^2

Simplifying:

\displaystyle mv^2=kx^2

Dividing by k:

\displaystyle x^2=\frac{mv^2}{k}

Taking square root:

\displaystyle x=\sqrt{\frac{mv^2}{k}}=v\sqrt{\frac{m}{k}}

The distance that the spring compresses is \mathbf{v\sqrt{\frac{m}{k}}}

Troyanec [42]3 years ago
7 0

Answer:

Explanation:

KE=PE

1/2mv^2=1/2kx^2

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

MV^2=kX^2

(MV^2)/k=X^2

X=√(mv^2)/k

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