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).
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(b) How many cm down the plane will block 1 have traveled when 0.47 s has elapsed?
To solve this problem, we can use Newton's laws of motion and the equations of motion. Let's first solve part (a) to find the magnitude of the acceleration of block 1 when it is released.
(a) Find the magnitude of the acceleration of block 1 when it is released (in m/s2):
To find the acceleration of block 1, we need to consider the forces acting on it. The forces involved are gravity (mg), the normal force (N), and kinetic friction (fk). The normal force can be decomposed into its vertical and horizontal components, N perpendicular to the inclined plane and N parallel to the inclined plane.
The acceleration can be determined by resolving the forces parallel to the inclined plane:
Net force parallel to the inclined plane, F = m1 * a
Net force parallel to the inclined plane, F = mg * sin(θ) - fk
The gravitational force component parallel to the inclined plane is mg * sin(θ).
The frictional force fk is given by fk = μ * N, where μ is the coefficient of kinetic friction.
To calculate the normal force component parallel to the inclined plane, we can use N = m1 * g * cos(θ).
Now, we can substitute the values and calculate the acceleration:
F = m1 * a
mg * sin(θ) - μ * N = m1 * a
m1 * g * sin(θ) - μ * m1 * g * cos(θ) = m1 * a
Substituting the given values:
m1 = 28 kg
θ = 47 degrees
μ = 0.8
g = 9.81 m/s^2
a = (28 * 9.81 * sin(47) - 0.8 * 28 * 9.81 * cos(47)) / 28
a ≈ 1.695 m/s^2
Therefore, the magnitude of the acceleration of block 1 when it is released is approximately 1.695 m/s^2.
Now, let's move on to part (b) to find how many cm down the plane block 1 will have traveled when 0.47 s has elapsed.
(b) How many cm down the plane will block 1 have traveled when 0.47 s has elapsed:
To solve this part, we can use the equation of motion:
S = ut + (1/2) * a * t^2
where S is the distance traveled, u is the initial velocity (which is 0 since the system is released from rest), a is the acceleration, and t is the time.
Given:
t = 0.47 s
a = 1.695 m/s^2
Substituting the values:
S = 0 * 0.47 + (1/2) * 1.695 * (0.47)^2
S ≈ 0.1846 m
To convert this to cm, we multiply by 100:
S ≈ 0.1846 * 100
S ≈ 18.46 cm
Therefore, when 0.47 s has elapsed, block 1 will have traveled approximately 18.46 cm down the plane.