Question

You give a crate of mass m=1.00kg an initial speed of vi=4.00m/s up an incline of θ=45.7. The crate then slides along the incline, reaches a spring of spring constant k=68.9N/m, and compresses the spring by Δs=15.0cm before stopping.

a. assuming there is no friction, what is the distance, d, in meters, along the incline that the crate slides?

b. what distance, d, in meters, had the crate moved when it just touched the spring?

c. if there were friction between the crate and incline, what is the equation for the magnitude of the kinetic friction force?

d. if there were friction between the crate and incline, what is the value of the coefficient of kinetic friction would stop the crate after it travels a distance, ds, stopping just before it begins to compress the spring?

Answer 1

a. The distance, d, along the incline that the crate slides is approximately 1.88 meters. b. The distance, d, that the crate had moved when it just touched the spring is approximately 0.154 meters. c. The magnitude of the kinetic friction force is given by f_k = μ_k * m * g * cos(theta), where μ_k is the coefficient of kinetic friction. d. The coefficient of kinetic friction that would stop the crate after it travels a distance, ds, stopping just before it begins to compress the spring is given by μ_k = (m * vi^2) / (2 * m * g * cos(theta) * ds).

Step-by-step explanation:

a. To determine the distance, d, along the incline that the crate slides, we can use the work-energy principle. The work done by the gravitational force on the crate is equal to the change in its gravitational potential energy. This can be calculated as m * g * d * sin(theta), where m is the mass of the crate, g is the acceleration due to gravity, and theta is the angle of the incline.

Equating this to the change in kinetic energy, we get m * vi^2 / 2. Solving for d, we find d = (m * vi^2)/(2 * m * g * sin(theta)). Plugging in the given values, we get d = (1.00 * 4.00^2) / (2 * 1.00 * 9.8 * sin(45.7)). The distance, d, along the incline is approximately 1.88 meters.

b. When the crate just touches the spring, it has come to rest. This means that its final kinetic energy is zero. Using the work-energy principle again, we can set the change in kinetic energy equal to the work done by the spring. This can be calculated as (1/2) * k * Δs^2, where k is the spring constant and Δs is the compression of the spring. Plugging in the given values, we get (1/2) * 68.9 * (0.15)^2. The distance, d, that the crate had moved when it just touched the spring is approximately 0.154 meters.

c. If there were friction between the crate and incline, the magnitude of the kinetic friction force can be calculated using the equation f_k = μ_k * m * g * cos(theta), where μ_k is the coefficient of kinetic friction, m is the mass of the crate, g is the acceleration due to gravity, and theta is the angle of the incline.

d. If there were friction between the crate and incline and we want to find the coefficient of kinetic friction that would stop the crate after it travels a distance, ds, stopping just before it begins to compress the spring, we can use the work-energy principle again. The work done by the friction force is equal to the change in kinetic energy. This can be calculated as f_k * ds, where f_k is the magnitude of the kinetic friction force and ds is the distance the crate travels.

Equating this to the change in kinetic energy (which is the final kinetic energy of the crate), we get f_k * ds = m * vi^2 / 2. Solving for μ_k, we find μ_k = (m * vi^2) / (2 * m * g * cos(theta) * ds). Plugging in the given values, we get μ_k = (1.00 * 4.00^2) / (2 * 1.00 * 9.8 * cos(45.7) * ds).