A block of mass m1 = 28 kg rests on a wedge of angle θ = 47∘ which is itself attached to a table (the wedge does not move in this problem). An inextensible string is attached to m1, passes over a frictionless pulley at the top of the wedge, and is then attached to another block of mass m2 = 3 kg. The coefficient of kinetic friction between block 1 and the plane is μ = 0.8. The string and wedge are long enough to ensure neither block hits the pulley or the table in this problem, and you may assume that block 1 never reaches the table. take g to be 9.81 m/s2.

The system is released from rest as shown above, at t = 0.

(a) Find the magnitude of the acceleration of block 1 when it is released (in m/s2).

(b) How many cm down the plane will block 1 have traveled when 0.47 s has elapsed?

To find the magnitude of the acceleration of block 1 when it is released, we can start by drawing a free-body diagram for each block and applying Newton's second law of motion.

(a) Free-body diagram and equations for block 1:

- The gravitational force acting on block 1 is given by F1_gravity = m1 * g * cos(θ), where g is the acceleration due to gravity and θ is the angle of the wedge.
- The normal force acting on block 1 is given by N1 = m1 * g * sin(θ), since the wedge supports the weight of block 1 perpendicular to the inclined plane.
- The force of friction acting on block 1 is given by F1_friction = μ * N1, where μ is the coefficient of kinetic friction.
- The net force on block 1 is given by F1_net = m1 * a1, where a1 is the acceleration of block 1.

Applying Newton's second law in the direction perpendicular to the inclined plane, we have:

N1 - F1_gravity * sin(θ) = 0

Simplifying, we get:

m1 * g * sin(θ) = m1 * g * cos(θ) * sin(θ)

Therefore, sin(θ) cancels out, and we have:

g = a1

So, the magnitude of the acceleration of block 1 when it is released is equal to the acceleration due to gravity, which is 9.81 m/s^2.

(b) To find how far block 1 travels down the plane in 0.47 s, we can use the equation of motion:

d = v0 * t + (1/2) * a * t^2

where d is the displacement, v0 is the initial velocity, t is the time, and a is the acceleration. In this case, the initial velocity is 0 m/s, the time is 0.47 s, and the acceleration is 9.81 m/s^2.

Plugging in the values, we get:

d = 0 * 0.47 + (1/2) * 9.81 * (0.47)^2

Simplifying, we find that block 1 will have traveled approximately 0.1035 meters or 10.35 cm down the plane when 0.47 s has elapsed.