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 can use Newton's second law and the principles of forces and acceleration. Here's how you can find the answers to the two parts of the question:
(a) Find the magnitude of the acceleration of block 1 when it is released (in m/s^2):
1. Determine the forces acting on block 1:
- The weight of block 1 (mg1) acts vertically downwards.
- The normal force (N) acts perpendicular to the inclined plane.
- The frictional force (f) acts parallel to the inclined plane, opposing motion.
2. Resolve the weight of block 1 into components:
- The component of the weight perpendicular to the inclined plane is mg1 * cos(θ).
- The component of the weight parallel to the inclined plane is mg1 * sin(θ).
3. Calculate the normal force:
- Since block 1 is not moving vertically, the vertical component of the weight is balanced by the normal force (N). So, N = mg1 * cos(θ).
4. Calculate the frictional force:
- The frictional force (f) is given by f = μ * N, where μ is the coefficient of kinetic friction.
5. Find the net force acting on block 1:
- The net force in the direction parallel to the inclined plane is given by F(net) = m1 * a, where a is the acceleration.
6. Write the equation of motion:
- F(net) = m1 * a = mg1 * sin(θ) - f.
7. Substitute the values:
- m1 = 28 kg, g = 9.81 m/s^2, θ = 53°, μ = 0.9.
8. Solve for the acceleration (a):
- m1 * a = m1 * g1 * sin(θ) - μ * N
- Substitute the values of m1, g1, θ, and μ.
- Solve for a.
(b) How many cm down the plane will block 1 have traveled when 0.50 s has elapsed:
To solve this part, we need to find the displacement of block 1 at t = 0.50 s.
1. Calculate the initial velocity of block 1 (u1):
- At t = 0, block 1 starts from rest, so u1 = 0 m/s.
2. Calculate the displacement (s1) using the equation of motion:
- s1 = u1 * t + (1/2) * a * t^2
- Substitute the values of u1, t, and a.
- Solve for s1.
Remember to convert the final answer to centimeters, as requested.
By following these steps, you should be able to find the answers to parts (a) and (b) of the question.