A block of mass m1 = 28 kg rests on a wedge of angle θ = 53∘ 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 = 4 kg. The coefficient of kinetic friction between block 1 and the plane is μ = 0.9. 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 0.50 s has elapsed?
To solve this problem, we will break it down into two parts: analyzing the forces acting on the blocks and applying Newton's laws of motion.
(a) Find the magnitude of the acceleration of block 1 when it is released (in m/s^2):
1. Identify the forces acting on block 1:
- Weight of block 1 (m1 * g) acting vertically downwards.
- Normal force (N1) exerted by the wedge acting perpendicular to the incline.
- Friction force (f1) opposing the motion, directed up the incline.
- Tension force (T) in the string pulling block 1.
2. Decompose the weight of block 1 into components:
- The component parallel to the incline is m1 * g * sin(θ).
- The component perpendicular to the incline is m1 * g * cos(θ).
3. Resolve the forces parallel and perpendicular to the incline:
- Parallel forces: T - f1 - m1 * g * sin(θ) = m1 * a (since block 1 moves only along the incline)
- Perpendicular forces: N1 - m1 * g * cos(θ) = 0 (as there is no vertical acceleration)
4. Solve for the friction force:
- The friction force f1 can be calculated using the equation f1 = μ * N1, where μ is the coefficient of kinetic friction between block 1 and the plane.
5. Substitute the values:
- The equation becomes T - μ * N1 - m1 * g * sin(θ) = m1 * a
- The normal force N1 can be calculated using the equation N1 = m1 * g * cos(θ)
6. Substitute the values of m1, g, θ, and μ into the equations and solve for the acceleration a.
(b) How many cm down the plane will block 1 have traveled when 0.50 s has elapsed:
1. Once the acceleration a is found, we can use the kinematic equation to find the displacement of block 1 after 0.50 s:
- s = ut + (1/2) at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
2. Calculate the initial velocity u:
- Since the system is released from rest, u = 0.
3. Substitute the values of a and t into the equation and solve for the displacement s.
Note: Make sure to convert the final displacement to centimeters, as required in the question.