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

`d'=√(2) d`

Step-by-step explanation:

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

`U=(kd^(2) )/(2)`

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

`K=(mv^(2) )/(2)`

By conservation of energy we have:

`(mv^(2) )/(2)=(kd^(2) )/(2)` (1)

If we double the kinetic energy

`2((mv^(2) )/(2))=(kd'^(2) )/(2)` (2)

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

`2((kd^(2) )/(2))=(kd'^(2) )/(2)`

`2((d^(2) )/(2))=(d'^(2) )/(2)`

`d'=√(2) d`

Answer 2

The compression is
`√(2) \  d`.

Step-by-step explanation:

A Hooke's law spring compressed has a potential energy

`E_(potential) = (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) = (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.

`(1)/(2) m v_1^2 = (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 * ((1)/(2) m v_1^2) = (1)/(2) k D^2`

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

`2 * ((1)/(2) k d^2) = (1)/(2) k D^2`

`D^2 =  (2 \ ((1)/(2) k d^2))/((1)/(2) k)`

`D^2 =  2 \  d^2`

`D =  √(2 \  d^2)`

`D =  √(2) \  d`

And this is the compression we are looking for