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?

m₁a=m₁gsinα-T-F(fr)… (1)

0=N-m₁gcosα………….(2)
m₂a=T-m₂g…………….(3)
F(fr)=μN=μm₁gcosα
sum of (1) and (3):
a(m₁+m₂)=m₁gsinα-T-F(fr)+ T-m₂g=
= m₁gsinα - μm₁gcosα - m₂g
a=g(m₁sinα - μm₁cosα - m₂)/(m₁+m₂) =…
s=at₂/2=….

To find the magnitude of the acceleration of block 1 when it is released, we need to analyze the forces acting on the system.

(a) Forces acting on block 1:
- Weight: The weight of block 1 acts vertically downward and has a magnitude of m1 * g, where g is the acceleration due to gravity. So, the weight of block 1 is 28 kg * 9.81 m/s^2 = 274.68 N.
- Normal Force: The normal force acts perpendicularly to the surface of the wedge. Since the wedge does not move, the normal force is equal in magnitude and opposite in direction to the component of the weight of block 1 perpendicular to the wedge's surface, which is m1 * g * cos(θ). So, the normal force is 28 kg * 9.81 m/s^2 * cos(47°) = 196.37 N.
- Friction Force: The friction force acts parallel to the surface of the wedge and opposes the motion of block 1. It has a magnitude equal to the coefficient of kinetic friction (μ) multiplied by the normal force. So, the friction force is 0.8 * 196.37 N = 157.10 N.

The net force acting on block 1 is the difference between the force parallel to the surface of the wedge (friction force) and the component of the weight parallel to the surface. It is given by:
Net Force = Friction Force - Component of Weight parallel to the surface
Net Force = 157.10 N - (m1 * g * sin(θ))
Net Force = 157.10 N - (28 kg * 9.81 m/s^2 * sin(47°))

We can now use Newton's Second Law (F = m * a) to find the acceleration of block 1:
m1 * a = Net Force
a = Net Force / m1
a = (157.10 N - (28 kg * 9.81 m/s^2 * sin(47°))) / 28 kg

Calculating this value will give us the magnitude of the acceleration of block 1 when it is released.

(b) To find how far block 1 has traveled down the plane after a given time, we need to calculate the displacement. Using the equation of motion:
displacement = initial velocity * time + (1/2) * acceleration * time^2

Since the system is released from rest, the initial velocity is zero:
displacement = (1/2) * acceleration * time^2

Substituting the known values, acceleration from part (a), and the given time of 0.47 s will give us the distance traveled by block 1 down the plane.