Question

A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface. What compression of the spring would result in the mass attaining double the kinetic energy received in the above situation?

Answer 1

Answer:

The compression is $\sqrt{2} \  d$.

Explanation:

A Hooke's law spring compressed has a potential energy

$E_{potential} = \frac{1}{2} k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_{kinetic} = \frac{1}{2} m v^2$.

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$. Knowing that the energy is constant.

$\frac{1}{2} m v_1^2 = \frac{1}{2} k d^2$

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

$2 * (\frac{1}{2} m v_1^2) = \frac{1}{2} k D^2$

But, in the left side we can use the previous equation to obtain:

$2 * (\frac{1}{2} k d^2) = \frac{1}{2} k D^2$

$ D^2 =  \frac{2 \ (\frac{1}{2} k d^2)}{\frac{1}{2} k}$

$ D^2 =  2 \  d^2$

$ D =  \sqrt{2 \  d^2}$

$ D =  \sqrt{2} \  d$

And this is the compression we are looking for

Answer 2

Answer:

$d'=\sqrt{2} d$

Explanation:

By hooke's law we have that the potential energy can be defined as:

$U=\frac{kd^{2} }{2}$

Where k is the spring constant and d is the compression distance, the kinetic energy can be written as

$K=\frac{mv^{2} }{2}$

By conservation of energy we have:

$\frac{mv^{2} }{2}=\frac{kd^{2} }{2}$ (1)

If we double the kinetic energy

$2(\frac{mv^{2} }{2})=\frac{kd'^{2} }{2}$ (2)

where d' is the new compression, now if we input (1) in (2) we have

$2(\frac{kd^{2} }{2})=\frac{kd'^{2} }{2}$

$2(\frac{d^{2} }{2})=\frac{d'^{2} }{2}$

$d'=\sqrt{2} d$