A light spring having a force constat of 125 N/m is used to pull a 9.50kg sled on a horizontal frixtionless ice rink. If the sled has an acceleration of 2.00 m/s^2, by how much does the spring stretch if it pulls on the sled (a) horizontally, (b) at 30.0 degrees above the horizontal

For b i don't understand how to get the answer they got and that is 17.6 cm
If at 30 deg above.
x= ma/k (1/cos30)

But then theres a second part to the question asking what would the answers in both cases be if there were friction and the coefficient of knetic friction between the sled and the ice were .200?

(b) The spring must supply a force F uch tham
F - Mg sin 30 = M a

The term on the left is the net force up the incline, and has the component of weight subtracted. Solve for F , then use the same spring formula for the deflection.

M g sin 30 = 46.55 N
F = 46.55 + 19 = 65.55 N
deflection = 65.55/125 = 0.524 m

With friction, subtract the friction force, which you can calculate, from the applied force; then apply
Fnet = M a

the answer for b is suppose to be in cm and they got 17.6cm

and for the second part of the question about adding force would it be the 125N - .200 ??

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.