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?
all i know is how to find acceleration. formula is (f - mu_1*m_b*g - mu_2*m_s*g)/m_b
To find the magnitude of the acceleration of block 1 when it is released, we can first analyze the forces acting on it.
1. Weight: The weight of block 1 acts vertically downward with a magnitude of m1 * g, where g = 9.81 m/s^2 is the acceleration due to gravity.
2. Normal Force: The normal force acts perpendicular to the inclined plane. As block 1 is at rest, the normal force equals the component of the weight of block 1 perpendicular to the plane, which is m1 * g * cos(θ), where θ = 47° is the angle of the wedge.
3. Friction Force: The friction force acts parallel to the inclined plane and opposes the motion of block 1. The magnitude of the friction force can be calculated as μ * (normal force), where μ = 0.8 is the coefficient of kinetic friction. Therefore, the friction force is 0.8 * m1 * g * cos(θ).
4. Tension Force: The tension force in the string acts upward and is the force responsible for accelerating both blocks. Since the string is inextensible, the tension is the same throughout the length of the string.
Next, we apply Newton's second law of motion to block 1 in the direction parallel to the inclined plane:
m1 * a = T - friction force
Substituting the expressions for the friction force and the tension force:
m1 * a = T - 0.8 * m1 * g * cos(θ)
As block 1 is attached to block 2 by the string, the tensions in the string on both sides of the pulley are equal. Thus, the tension force can be expressed in terms of block 2:
T = m2 * g
Substituting this into the equation:
m1 * a = m2 * g - 0.8 * m1 * g * cos(θ)
Dividing both sides by m1:
a = (m2 * g - 0.8 * m1 * g * cos(θ)) / m1
Plug in the given values:
m1 = 28 kg
m2 = 3 kg
g = 9.81 m/s^2
θ = 47°
a = (3 kg * 9.81 m/s^2 - 0.8 * 28 kg * 9.81 m/s^2 * cos(47°)) / 28 kg
Calculate this expression to find the magnitude of the acceleration of block 1 when it is released.
For part (b), we can use the kinematic equation to determine how far block 1 has traveled down the plane after 0.47 s. The kinematic equation for linear motion with constant acceleration is:
s = ut + (1/2) * a * t^2
Where:
s is the distance traveled
u is the initial velocity (which is 0 as block 1 is released from rest)
a is the acceleration
t is the time elapsed (0.47 s in this case)
To find the distance traveled, substitute the values into the equation:
s = (1/2) * a * t^2
Plug in the calculated value for acceleration and the given value of time:
s = (1/2) * (magnitude of acceleration) * (0.47 s)^2
Calculate this expression to find how many cm down the plane block 1 will have traveled after 0.47 s. Remember to convert the units if necessary.