Answer:

Explanation:
Here we will call:
1.
: The energy when the first spring is compress
2.
: The energy after the mass is liberated by the spring
3.
: The energy before the second string catch the mass
4.
: The energy when the second sping compressed
so, the law of the conservations of energy says that:
1. 
2. 
3.
where
is the work of the friction.
1. equation 1 is equal to:
where K is the constant of the spring, x is the distance compressed, M is the mass and
the velocity, so:
Solving for velocity, we get:
= 65.319 m/s
2. Now, equation 2 is equal to:
where M is the mass,
the velocity in the situation 2,
is the velocity in the situation 3,
is the coefficient of the friction, N the normal force and d the distance, so:
Volving for
, we get:

3. Finally, equation 3 is equal to:

where
is the constant of the second spring and
is the compress of the second spring, so:

solving for
, we get:
