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 use Newton's second law of motion. The force acting on block 1 is the force of gravity and the force of kinetic friction. The force of gravity acting on block 1 can be calculated using the formula: F_gravity = m1 * g, where m1 is the mass of block 1 and g is the acceleration due to gravity.
The force of kinetic friction can be calculated using the formula: F_friction = μ * (m1 * g * cos(θ)), where μ is the coefficient of kinetic friction and θ is the angle of the wedge.
The net force acting on block 1 is the difference between the force of gravity and the force of kinetic friction: F_net = F_gravity - F_friction.
Now, applying Newton's second law of motion: F_net = m1 * a, where a is the acceleration of block 1.
Substituting the values in the previous equations and solving for a, we get:
m1 * a = m1 * g - μ * (m1 * g * cos(θ))
Dividing both sides by m1, we get:
a = g - μ * (g * cos(θ))
Substituting the given values for g, μ, and θ, we get:
a = 9.81 - 0.8 * (9.81 * cos(47))
Calculating this value will give us the answer to part (a).
To find how many cm down the plane block 1 will have traveled when 0.47 s has elapsed, we can use the equations of motion. The equation that relates the distance traveled, initial velocity, time, and acceleration is:
s = ut + (1/2) * a * t^2
Where s is the distance traveled, u is the initial velocity (which is 0 in this case, as the block is released from rest), t is the time, and a is the acceleration.
Given that the time elapsed is 0.47 s and we have already calculated the acceleration in part (a), we can substitute these values into the equation and calculate the distance traveled. Since the question asks for the distance in cm, we need to convert the answer to cm by multiplying it by 100.
Calculating this value will give us the answer to part (b).