A crate of mass 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 1140 N/m. The mass is pressed against the spring a
distance x and released. It moves up the slope
0.154 meters from the compressed position before
coming to a stop. If the cofficient of kinetic friction is 0.195,
how far (m) was the spring compressed?
To calculate how far the spring was compressed, we need to use the laws of physics related to the forces acting on the crate.
First, let's identify the forces acting on the crate:
1. Gravitational force (mg): This force acts straight down the incline and can be calculated using the mass of the crate (m = 0.812 kg) and the acceleration due to gravity (9.8 m/s^2).
Gravitational force (mg) = 0.812 kg × 9.8 m/s^2
2. Normal force (N): This force acts perpendicular to the incline and balances the component of gravity perpendicular to the incline. It can be calculated by taking the vertical component of the gravitational force.
Normal force (N) = mg × cos(θ)
3. Friction force (Ffriction): This force acts parallel to the incline and opposes the crate's motion. It can be calculated using the coefficient of kinetic friction (μ = 0.195) and the normal force (N).
Friction force (Ffriction) = μ × N
Now that we have identified the forces, we can analyze the motion of the crate using Newton's second law.
The component of the gravitational force parallel to the incline (mg × sin(θ)) is balanced by the friction force (Ffriction). Therefore, we can set up the equation:
mg × sin(θ) = Ffriction
Substituting the expressions for each force, we get:
0.812 kg × 9.8 m/s^2 × sin(35.3°) = 0.195 × mg × cos(35.3°)
Now, let's solve for mg × cos(35.3°):
mg × cos(35.3°) = (0.812 kg × 9.8 m/s^2 × sin(35.3°)) / 0.195
Now, we need to find the compression distance x at which the spring force (Fspring) balances the component of the gravitational force along the incline.
Using Hooke's law, the spring force (Fspring) can be calculated by multiplying the spring constant (k = 1140 N/m) by the compression distance x.
Fspring = k × x
The component of the gravitational force along the incline (mg × sin(θ)) is balanced by the spring force (Fspring). Therefore, we can set up the equation:
Fspring = mg × sin(θ)
Substituting the expressions for each force and the previously calculated value of mg × cos(35.3°), we get:
k × x = (0.812 kg × 9.8 m/s^2 × sin(35.3°)) / 0.195
Now, let's solve for x:
x = [(0.812 kg × 9.8 m/s^2 × sin(35.3°)) / 0.195] / k
Evaluating this expression will give us the distance x by which the spring was compressed.