Three blocks are in contact with each other on a frictionless, horizontal surface, as shown below. A horizontal force is applied to m1. Take m1 = 2.00 kg, m2 = 3.00 kg, m3 = 5.00 kg, and F = 22.0 N
To solve this problem, we can use Newton's laws of motion and apply them to each block individually.
Let's start by analyzing Block 1 (m1). We know that a horizontal force (F) is applied to this block. Since the surface is frictionless, the only force acting on this block is the applied force. Using Newton's second law (F = ma), we can calculate the acceleration of Block 1.
F = ma1
22.0 N = (2.00 kg) * a1
a1 = 11.0 m/s^2
Now, let's move to Block 2 (m2). Block 2 is in contact with Block 1, so it experiences the same acceleration as Block 1. Here, we consider the net force on Block 2, which is due to the force applied to Block 1.
Fnet = m2 * a2
Fnet = (3.00 kg) * (11.0 m/s^2)
Fnet = 33.0 N
The net force acting on Block 2 is 33.0 N.
Finally, let's analyze Block 3 (m3). Similar to Block 2, Block 3 also experiences the same acceleration as Blocks 1 and 2 since it is in contact with Block 2.
Fnet = m3 * a3
Fnet = (5.00 kg) * (11.0 m/s^2)
Fnet = 55.0 N
The net force acting on Block 3 is 55.0 N.
To summarize:
- The acceleration of Block 1 (m1) is 11.0 m/s^2.
- The net force acting on Block 2 (m2) is 33.0 N.
- The net force acting on Block 3 (m3) is 55.0 N.
Please note that these calculations assume an idealized setup with no friction and neglects the weight of each block.