A light spring having a force constant of 110 N/m is used to pull a 10.0 kg sled on a horizontal frictionless ice rink. The sled has an acceleration of 2.0 m/s^2

(a)By how much does the spring stretch if it pulls on the sled horizontally?

(b)By how much does the spring stretch if it pulls on the sled at 30.0^\circ above the horizontal?

To solve this problem, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.

(a) By how much does the spring stretch if it pulls on the sled horizontally?

We can start by calculating the force exerted by the spring by multiplying the force constant (k) with the displacement (x). The force exerted by the spring is equal to the mass of the sled (m) multiplied by its acceleration (a). Therefore, we have:

Force exerted by the spring (F_spring) = k * x

Force exerted by the spring (F_spring) = m * a

Since these two forces are the same, we can set them equal to each other:

k * x = m * a

Rearranging the equation to solve for x, we have:

x = (m * a) / k

Substituting the given values, we get:

x = (10.0 kg * 2.0 m/s^2) / 110 N/m

x ≈ 0.182 m

Therefore, the spring stretches by approximately 0.182 meters when it pulls on the sled horizontally.

(b) By how much does the spring stretch if it pulls on the sled at 30.0 degrees above the horizontal?

In this case, we need to consider the component of the force exerted by the spring that acts horizontally. Let's call this force F_horizontal.

F_horizontal = F_spring * cos(30°)

Using the same equation as before, substituting F_horizontal for F_spring:

k * x = m * a * cos(30°)

Rearranging the equation to solve for x, we have:

x = (m * a * cos(30°)) / k

Substituting the given values, we get:

x = (10.0 kg * 2.0 m/s^2 * cos(30°)) / 110 N/m

x ≈ 0.151 m

Therefore, the spring stretches by approximately 0.151 meters when it pulls on the sled at 30.0 degrees above the horizontal.

To calculate the spring stretch, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the amount it stretches or compresses. The equation for Hooke's Law is given by:

F = -kx

Where F is the force applied by the spring, k is the spring constant, and x is the amount the spring stretches (or compresses).

(a) By how much does the spring stretch if it pulls on the sled horizontally?

In this case, the spring is pulling the sled horizontally. The force applied by the spring is given by the formula:

F = ma

Where m is the mass of the sled and a is the acceleration of the sled.

Given:
Force constant, k = 110 N/m
Mass of the sled, m = 10.0 kg
Acceleration of the sled, a = 2.0 m/s^2

Substituting these values into the equation, we have:

F = ma
110 N/m * x = 10.0 kg * 2.0 m/s^2

Simplifying the equation:

110 N/m * x = 20.0 kg·m/s^2

Dividing both sides by 110 N/m, we get:

x = 0.182 m

Therefore, the spring stretches by 0.182 meters when it pulls the sled horizontally.

(b) By how much does the spring stretch if it pulls on the sled at 30.0 degrees above the horizontal?

In this case, we need to calculate the component of the force applied by the spring in the horizontal direction. Since the spring is pulling at an angle, the horizontal component of the force is given by:

F_horizontal = F * cos(30°)

The force applied by the spring is still given by F = ma, so we can substitute it into the equation:

F_horizontal = ma * cos(30°)
110 N/m * x = 10.0 kg * 2.0 m/s^2 * cos(30°)

Simplifying the equation:

110 N/m * x = 20.0 kg·m/s^2 * cos(30°)

Dividing both sides by 110 N/m, we get:

x = 0.160 m

Therefore, the spring stretches by 0.160 meters when it pulls the sled at 30.0 degrees above the horizontal.