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?

A. 24 cm
B.360 cm
C.14 cm
D. 240 cm
E. 36 cm

How did you start this problem?

I do not know what formula to use

I really need some help on this one

To find out how much the spring stretches with the brick attached, we need to solve this problem using the concept of forces and equilibrium.

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

1. Weight (mg): The weight of the brick pulling it downward. Its magnitude can be calculated using the formula mg, where m is the mass of the brick and g is the acceleration due to gravity (approximately 9.8 m/s^2).

2. Normal force (N): The force exerted by the inclined plane perpendicular to the surface. It acts in the opposite direction to the weight and can be calculated using the formula N = mg * cosθ, where θ is the angle of the inclined plane (34° in this case).

3. Force due to the spring (F_spring): The force exerted by the spring on the brick. This force acts in the opposite direction to the normal force.

Since the brick is in equilibrium (not moving), the sum of the forces acting on it must be zero. Therefore, the magnitude of the force due to the spring must be equal to the magnitude of the weight.

Let's calculate the magnitudes of the weight and normal force:

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

Normal force (N) = (3.0 kg) * (9.8 m/s^2) * cos(34°) ≈ 24.6 N

Since the force due to the spring (F_spring) must be equal to the weight (29.4 N), we can use Hooke's Law to calculate the amount of spring stretch (x):

F_spring = k * x

where k is the spring constant (force constant) and x is the amount of spring stretch.

Therefore, x = F_spring / k

Substituting the values, we get:

x = 29.4 N / (120 N/m) ≈ 0.245 m

To convert the value from meters to centimeters, we multiply by 100:

x ≈ 0.245 m * 100 cm/m ≈ 24.5 cm

The spring stretches approximately 24.5 cm.

Since this value is closest to option A, the correct answer is A. 24 cm.