Ahooke'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?

The compression is .

Explanation:

A Hooke's law spring compressed has a potential energy

where k is the spring constant and the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

.

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

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

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

And this is the compression we are looking for

Explanation:

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

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

By conservation of energy we have:

(1)

If we double the kinetic energy

(2)

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

i think they are all correct

explanation:

they seem good

c. newton's third law of motion