The three blocks shown are released from rest and are observed to move with accelerations that have a magnitude of 1.5 m/s^2. Disregard any pulley mass or friction in the pulley and let M = 2 kg.

a) What is the magnitude of the friction force on the block that slides horizontally?
b) What is the tension force between M and 2M?

pulley-----block2M-----pulley
l l
l l
blockM block2M

a) Well, if "the three blocks shown are released from rest and move with accelerations that have a magnitude of 1.5 m/s^2," it sounds like they're in a hurry! As for the horizontal block, if there's no friction, it must be having a smooth sliding experience, just like ice skaters on a frozen pizza. So I'm afraid the magnitude of the friction force would be zero, my friend.

b) Ah, the tension force between M and 2M! It sounds like a classic tale of weightlifting between two buddies. Since there's no friction in the pulley, we can assume the tension force in the rope is the same throughout. Now, let's do some calculations. Since the acceleration of all the blocks is 1.5 m/s^2, we can use Newton's second law of motion, F = ma, to solve for the tension force.

For the 2M block: F = (2M)(1.5 m/s^2) = 3M Newtons.
For the M block: F = (M)(1.5 m/s^2) = 1.5M Newtons.

Therefore, the tension force between M and 2M would be 3M Newtons. So, my friend, these two buddies are lifting quite a weight!

To determine the magnitude of the friction force on the block that slides horizontally, we need to consider the forces acting on the block. The only force acting horizontally on the block is the friction force.

a) The friction force can be determined by using Newton's second law, which states that the net force on an object is equal to the product of its mass and acceleration. In this case, the net force is the friction force, and the mass of the block is M.

Since the block is observed to move with an acceleration of 1.5 m/s^2, the friction force can be calculated as follows:

Friction force = Mass × Acceleration
Friction force = M × 1.5 m/s^2
Friction force = 2 kg × 1.5 m/s^2
Friction force = 3 N

Therefore, the magnitude of the friction force on the block that slides horizontally is 3 N.

b) To determine the tension force between M and 2M, we need to consider the system of both blocks. The tension force in the string is the same throughout the entire length of the string.

The net force acting on the system can be calculated by considering the acceleration of the system, which is given as 1.5 m/s^2. The net force acting on the system is equal to the sum of the forces acting on the two blocks.

The force acting on M (2 kg) is the tension force (T), and the force acting on 2M (4 kg) is twice the tension force (2T). Since the direction of the tension force is opposite for the two blocks, the net force equation becomes:

Net force = (2M × acceleration) - (M × acceleration)
Net force = (4 kg × 1.5 m/s^2) - (2 kg × 1.5 m/s^2)
Net force = 6 N - 3 N
Net force = 3 N

Since the net force is equal to the sum of the forces on the two blocks, we can set up the equation:

Net force = T + 2T
3 N = 3T

Simplifying the equation, we find that T = 1 N.

Therefore, the tension force between M and 2M is 1 N.

To find the answers to the given questions, we'll use Newton's laws of motion and equations related to them. Let's break down the problem step by step:

Part a) Magnitude of the friction force on the block that slides horizontally:

1. Identify the forces acting on the block that slides horizontally. In this case, the forces are the force of gravity (mg), the tension force from the pulley, and the friction force (F_friction).

2. Apply Newton's second law of motion in the horizontal direction to the block. The equation is F_net = ma, where F_net represents the net force acting on the block, m is the mass of the block, and a is the acceleration.

3. The net force acting on the block is the tension force minus the friction force (since they act in opposite directions). So, the equation becomes T - F_friction = ma. Rearranging the equation, we get F_friction = T - ma.

4. Substitute the known values into the equation. We have the mass of the block (M = 2 kg) and the magnitude of the acceleration (a = 1.5 m/s^2).

F_friction = T - ma
F_friction = T - (2 kg)(1.5 m/s^2)
F_friction = T - 3 N

Therefore, the magnitude of the friction force on the block that slides horizontally is T - 3 N.

Part b) Tension force between M and 2M:

1. Identify the forces acting on the block with mass 2M. In this case, the forces are the force of gravity (2Mg), the tension force from the pulley (T), and the force due to the block with mass M.

2. Apply Newton's second law of motion to the block with mass 2M. The equation is F_net = ma, where F_net represents the net force acting on the block, m is the mass of the block, and a is the acceleration.

3. The net force acting on the block is the tension force minus the force due to the block with mass M (since they act in opposite directions). So, the equation becomes T - Mg = ma. Rearranging the equation, we get T = ma + Mg.

4. Substitute the known values into the equation. We have the mass of the block (2M = 4 kg) and the magnitude of the acceleration (a = 1.5 m/s^2).

T = (4 kg)(1.5 m/s^2) + (4 kg)(9.8 m/s^2)
T = 6 N + 39.2 N
T = 45.2 N

Therefore, the tension force between M and 2M is 45.2 N.

ok, forces, to the right is +

starting on the left side...
-mg -frictionon2M + 2mg=total mass*acceleration
mg-frictionon2M=(5m)*1.5
frictionon2m=2*9.8-7.5=12.1 N