Question

a block of mass 3 kg slides down an inclined plane at an angle of p with a massless tether attached to a pulley with mass 1 kg and radius 0.5 m at the top of the incline (see the following figure). the pulley can be approximated as a disk. the coefficient of kinetic friction on the plane is 0.4. what is the acceleration of the block?

Answer 1

The acceleration of the block sliding down the inclined plane is 6.076 m/s^2.

Step-by-step explanation:

To find the acceleration of the block sliding down the inclined plane, we can use Newton's second law. The net force acting on the block is the difference between the force of gravity down the incline and the force of friction. The force of gravity down the incline is given by mg sin(theta), where m is the mass of the block and g is the acceleration due to gravity. The force of friction is given by mu_k mg cos(theta), where mu_k is the coefficient of kinetic friction. Therefore, the net force is given by:

F_net = mg sin(theta) - mu_k mg cos(theta)

Using the formula F_net = ma, where a is the acceleration, we can solve for the acceleration:

a = F_net / m = (mg sin(theta) - mu_k mg cos(theta)) / m

We can plug in the values given in the problem: m = 3 kg, theta = 45 degrees, and mu_k = 0.4. Substituting these values, we get:

a = (3kg * 9.8m/s^2 * sin(45) - 0.4 * 3kg * 9.8m/s^2 * cos(45)) / 3kg

a = (44.1 N - 25.872 N) / 3kg

a = 18.228 N / 3kg

a = 6.076 m/s^2