Question

A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface of length 2 m with a coefficient of friction 0.16, then compresses the second spring of constant 2 N/m. The acceleration of gravity is 9.8 m/s2.
How far will the second spring compress in order to bring the mass to a stop?

Answer 1

Answer:

$X_2=25.27m$

Explanation:

Here we will call:

1. $E_1$: The energy when the first spring is compress

2. $E_2$: The energy after the mass is liberated by the spring

3. $E_3$: The energy before the second string catch the mass

4. $E_4$: The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. $E_1 = E_2$

2. $E_2 -E_3= W_f$

3.$E_3 = E_4$

where $W_f$ is the work of the friction.

1. equation 1 is equal to:

$\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2$

where K is the constant of the spring, x is the distance compressed, M is the mass and $V_2$ the velocity, so:

$\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2$

Solving for velocity, we get:

$V_2$ = 65.319 m/s

2. Now, equation 2 is equal to:

$\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd$

where M is the mass, $V_2$ the velocity in the situation 2, $V_3$ is the velocity in the situation 3, $U_k$ is the coefficient of the friction, N the normal force and d the distance, so:

$\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2)$

Volving for $V_3$, we get:

$V_3 = 65.27 m/s$

3. Finally, equation 3 is equal to:

$\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2$

where $K_2$ is the constant of the second spring and $X_2$ is the compress of the second spring, so:

$\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2$

solving for $X_2$, we get:

$X_2=25.27m$