Respuesta :
Answer:
[tex]X_2=25.27m[/tex]
Explanation:
Here we will call:
1. [tex]E_1[/tex]: The energy when the first spring is compress
2. [tex]E_2[/tex]: The energy after the mass is liberated by the spring
3. [tex]E_3[/tex]: The energy before the second string catch the mass
4. [tex]E_4[/tex]: The energy when the second sping compressed
so, the law of the conservations of energy says that:
1. [tex]E_1 = E_2[/tex]
2. [tex]E_2 -E_3= W_f[/tex]
3.[tex]E_3 = E_4[/tex]
where [tex]W_f[/tex] is the work of the friction.
1. equation 1 is equal to:
[tex]\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2[/tex]
where K is the constant of the spring, x is the distance compressed, M is the mass and [tex]V_2[/tex] the velocity, so:
[tex]\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2[/tex]
Solving for velocity, we get:
[tex]V_2[/tex] = 65.319 m/s
2. Now, equation 2 is equal to:
[tex]\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd [/tex]
where M is the mass, [tex]V_2[/tex] the velocity in the situation 2, [tex]V_3[/tex] is the velocity in the situation 3, [tex]U_k[/tex] is the coefficient of the friction, N the normal force and d the distance, so:
[tex]\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2) [/tex]
Volving for [tex]V_3[/tex], we get:
[tex]V_3 = 65.27 m/s[/tex]
3. Finally, equation 3 is equal to:
[tex]\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2[/tex]
where [tex]K_2[/tex] is the constant of the second spring and [tex]X_2[/tex] is the compress of the second spring, so:
[tex]\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2[/tex]
solving for [tex]X_2[/tex], we get:
[tex]X_2=25.27m[/tex]
