Question

A crate of mass m = 0.812 kg is placed on a rough incline of angle 35.3 degrees. near the base of the incline is a spring of spring constant k = 1140 n/m. the mass is pressed against the spring so that the spring is compressed a distance x from its relaxed length and then released. the mass is not connected to the spring, so when the spring reaches its relaxed length, it does not pull back on the mass. instead the mass slides freely up the slope before coming to a stop. the total distance d the mass moves (from the compressed point to the final point) is d = 0.168m. if the coefficient of kinetic friction is uk = 0.195, how far was the spring compressed

Answer 1

Using the given values, the spring was compressed approximately 3.3 cm.

Step-by-step explanation:

To find how far the spring was compressed, we need to analyze the forces acting on the crate and apply Newton's laws of motion.

1. First, let's consider the forces acting on the crate on the incline:

• - The gravitational force (mg) acting vertically downward.
• - The normal force (N) acting perpendicular to the incline.
• - The force of kinetic friction (f`k) acting parallel to the incline.

2. The normal force can be calculated as:

N = mg * cos(θ)

where θ is the angle of the incline (35.3 degrees).

3. The force of kinetic friction can be calculated as:

f`k = uk * N

where uk is the coefficient of kinetic friction (0.195).

4. The net force acting on the crate can be calculated as:

F-net = mg * sin(θ) - f`k

where θ is the angle of the incline (35.3 degrees).

5. The net force is also equal to the force exerted by the spring:

F-spring = -k * x

where k is the spring constant (1140 N/m) and x is the distance the spring is compressed.

6. Setting the net force equal to the force exerted by the spring, we have:

mg * sin(θ) - u`k * mg * cos(θ) = -k * x

7. Rearranging the equation, we can solve for x:

x = (mg * sin(θ) - u`k * mg * cos(θ)) / k

8. Plugging in the given values:

• m = 0.812 kg
• θ = 35.3 degrees
• u`k = 0.195
• k = 1140 N/m

9. Now we can calculate x:

`x = (0.812 * 9.8 * sin(35.3) - 0.195 * 0.812 * 9.8 * cos(35.3)) / 1140`

10. Simplifying the equation, we find that x is approximately:

x ≈ 0.033 m or 3.3 cm

Therefore, the spring was compressed approximately 3.3 cm.