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.

To find how much the spring stretches, we need to determine the force applied by the spring and equate it to the weight component acting down the ramp.

Let's break down the forces acting on the brick:

1. Weight (mg): The weight of the brick is given by W = mg, where m represents the mass (3.0 kg) and g is the acceleration due to gravity (9.8 m/s^2). Therefore, the weight of the brick is W = 3.0 kg × 9.8 m/s^2 = 29.4 N.

2. Normal force (N): The normal force acts perpendicular to the surface of the ramp and balances the weight component acting down the ramp. Since the ramp is perfectly smooth, the normal force is equal in magnitude and opposite in direction to the weight of the brick. So, the normal force is N = -29.4 N.

3. Force applied by the spring (F_spring): This is the force that prevents the brick from sliding down the ramp. It acts in the direction opposite to the weight component along the ramp. To find this force, we need to use trigonometry.

The weight component acting down the ramp is given by W_ramp = W × sin(theta), where theta represents the angle of inclination (34 degrees). Therefore, W_ramp = 29.4 N × sin(34°) ≈ 16.35 N.

Since the spring force should balance the weight component, we have F_spring = -W_ramp = -16.35 N.

Now, we can use Hooke's Law to find the displacement of the spring:

F_spring = k × x

Where F_spring is the force applied by the spring, k is the spring constant (force constant), and x is the displacement of the spring.

Substituting the values, -16.35 N = 120 N/m × x.

Solving for x, we find x = -16.35 N / 120 N/m ≈ -0.14 m.

Therefore, the spring stretches by approximately 0.14 meters with the brick attached.