A 3.0 kg brick rests on a perfectly smooth ramp inclined at 34 above the horizontal. The brick is kept from sliding down the plane by an ideal spring that is aligned with the surface and attached to a wall above the brick. The spring has a spring constant (force constant) of 120 N/m. By how much does the spring stretch with the brick attached?

See previous post: Tue, 10-28-14, 12:16 AM.

To find the amount by which the spring stretches, we need to analyze the forces acting on the brick.

Let's break down the forces into components parallel and perpendicular to the incline.

1. Normal Force (perpendicular to the incline):

The normal force (N) balances out the component of the weight that is perpendicular to the incline. The weight of the brick (mg) can be calculated by multiplying the mass (m = 3.0 kg) by the acceleration due to gravity (g = 9.8 m/s^2).

Weight (mg) = 3.0 kg * 9.8 m/s^2 = 29.4 N

Since the ramp is smooth and there is no vertical acceleration, the normal force is equal in magnitude, but opposite in direction to the weight. Therefore, the normal force is 29.4 N.

2. Force of Gravity (parallel to the incline):

The force of gravity (mg) can be resolved into components parallel and perpendicular to the incline.

Force of gravity parallel to incline = mg * sin(θ)
Force of gravity perpendicular to incline = mg * cos(θ)

θ = 34 degrees (the angle of inclination)

Force of gravity parallel = 3.0 kg * 9.8 m/s^2 * sin(34°) = 16.09 N

3. Spring Force:

The spring force is equal in magnitude and opposite in direction to the component of the force of gravity that is parallel to the incline. Therefore, the spring force is 16.09 N.

According to Hooke's Law, the force exerted by the spring (F) is given by:

F = k * x

Where:
F = Force exerted by the spring (in N)
k = Spring constant (Force constant) = 120 N/m
x = Displacement or stretch of the spring (in m)

We can rearrange this equation to find the stretch of the spring (x):

x = F / k

Plugging in the values we have:

x = 16.09 N / 120 N/m ≈ 0.134 m (rounded to three decimal places)

Therefore, the spring stretches by approximately 0.134 meters when the brick is attached.

To find the amount by which the spring stretches with the brick attached, we can use the concept of equilibrium. In this case, the force due to gravity acting on the brick down the inclined plane is balanced by the force exerted by the spring.

Here's how you can approach the problem step by step:

1. Break down the weight of the brick into its components:
- The component of gravity acting parallel to the incline is mg sinθ, where m is the mass of the brick (3.0 kg) and θ is the angle of the incline (34 degrees).
- The component of gravity acting perpendicular to the incline is mg cosθ.

2. Determine the force exerted by the spring:
- The force exerted by the spring can be represented by Hooke's Law: F = kx, where F is the force exerted by the spring, k is the spring constant (120 N/m), and x is the amount of stretch or compression.

3. Equate the force components:
- Set the force due to gravity acting parallel to the incline equal to the force exerted by the spring: mg sinθ = kx.

4. Solve for x:
- Substitute the given values: (3.0 kg)(9.8 m/s^2)(sin(34 degrees)) = (120 N/m) x.
- Simplify and solve for x: x = [(3.0 kg)(9.8 m/s^2)(sin(34 degrees))] / (120 N/m).

By calculating the above expression, you should find that the spring stretches by approximately 0.3824 meters (or 38.24 cm) when the brick is attached to it on the inclined plane.