To find the answer to part (b) of the question, we need to break it down step by step.
1. First, let's consider the forces acting on the sled when the spring pulls it at a 30-degree angle above the horizontal:
- The weight of the sled, which can be broken down into two components: mg sin(30) acting down the incline, and mg cos(30) acting perpendicular to the incline.
- The tension force from the spring, acting horizontally.
2. To find the acceleration of the sled, we use Newton's second law: Fnet = ma.
- The net force acting on the sled is the tension force from the spring minus the component of the weight down the incline: Fnet = T - mg sin(30).
- Substituting the given values into the equation: Fnet = (125 N/m)x - (9.50 kg)(9.8 m/s^2)(sin(30)).
- Since the sled has an acceleration of 2.00 m/s^2, we can set Fnet equal to ma: 2.00 = (125 N/m)x - (9.50 kg)(9.8 m/s^2)(sin(30)).
3. Now, let's solve for x, which represents the stretch of the spring:
- Rearranging the equation: (125 N/m)x = 19 (9.8 m/s^2)(sin(30)) + (9.50 kg)(2.00 m/s^2).
- Simplifying: x = [(19 (9.8 m/s^2)(sin(30)) + (9.50 kg)(2.00 m/s^2)] / (125 N/m).
- Evaluating the expression: x ≈ 0.524 m.
Therefore, the answer to part (b) is 0.524 meters, which is equivalent to 52.4 centimeters.
For the second part of the question, if there is friction with a coefficient of kinetic friction of 0.200, the force of friction can be calculated using the equation: f = μN, where μ is the coefficient of kinetic friction and N is the normal force.
Assuming the sled is on a horizontal surface, the normal force N is equal to the weight of the sled: N = mg.
To find the net force acting on the sled, we subtract the force of friction from the applied force.
Using the equation Fnet = ma, we have:
Fnet = T - μN,
ma = T - μN.
Applying the spring formula for deflection, we can solve for x:
x = T / k,
x = (ma) / k.
Substituting the known values, we have:
x = (ma) / (125 N/m).
To find the deflection x, we need to calculate the force T:
T = mg sin(30) + f,
T = (9.50 kg)(9.8 m/s^2)(sin(30)) + (0.200)(9.50 kg)(9.8 m/s^2),
T = 46.55 N + 19 N,
T = 65.55 N.
Substituting the values back into the deflection equation:
x = (65.55 N)(2.00 m/s^2) / (125 N/m),
x ≈ 1.048 m.
Therefore, the answer to the second part of the question is approximately 1.048 meters, which is equivalent to 104.8 centimeters.