Question

A block of mass m is initially moving to the right on a horizontal frictionless surface at a
speed v. It then compresses a spring of spring constant k. At the instant when the kinetic energy
of the block is equal to the potential energy of the spring, the spring is compressed a distance of:

Answer 1

Answer:

The distance that the spring compresses is:

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

Explanation:

Kinetic and Elastic Potential Energy

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

Answer 2

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