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 need to analyze the forces acting on the block.
Let's start by drawing a free-body diagram for block 1:
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m1 |
______ ______|
| θ
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The forces acting on block 1 are:
1. The gravitational force (mg) pulling it downwards.
2. The normal force (N) exerted by the wedge perpendicular to the incline.
3. The friction force (f) opposing the motion of block 1.
4. The tension force (T) provided by the string.
The gravitational force can be split into two components:
1. The force down the incline (mg * sin(θ))
2. The force perpendicular to the incline (mg * cos(θ))
Since block 1 is not moving in the vertical direction, the normal force and the force perpendicular to the incline are equal in magnitude and cancel each other out.
The friction force (f) can be calculated using the formula:
f = μ * N
Since the force down the incline is greater than the force opposing it, we can write the equation of motion for block 1 along the incline as:
m1 * a = m1 * g * sin(θ) - f
Substituting the value of f:
m1 * a = m1 * g * sin(θ) - μ * N
We can express N in terms of m1, g, and θ:
N = m1 * g * cos(θ)
Substituting the value of N:
m1 * a = m1 * g * sin(θ) - μ * (m1 * g * cos(θ))
Simplifying the equation:
a = g * (sin(θ) - μ * cos(θ))
Now we can substitute the given values and calculate the answer to part (a).
Substituting θ = 47∘, μ = 0.8, and g = 9.81 m/s^2 into the equation:
a = 9.81 * (sin(47∘) - 0.8 * cos(47∘))
Calculating the value of sin(47∘) and cos(47∘):
sin(47∘) ≈ 0.7314
cos(47∘) ≈ 0.681
a ≈ 9.81 * (0.7314 - 0.8 * 0.681)
a ≈ 9.81 * (0.7314 - 0.5448)
a ≈ 9.81 * 0.1866
a ≈ 1.836 m/s^2
Therefore, the magnitude of the acceleration of block 1 when it is released is approximately 1.836 m/s^2.
Moving on to part (b), we need to calculate the distance traveled by block 1 down the incline after 0.47 s.
To find this, 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 m/s as the block is released from rest),
a is the acceleration, and
t is the time elapsed.
Substituting the given values into the equation:
s = 0 + (1/2) * 1.836 * (0.47)^2
Calculating the value:
s = 0.5 * 1.836 * 0.22^2
s ≈ 0.356 m
Therefore, after 0.47 s has elapsed, block 1 will have traveled approximately 0.356 meters down the incline.