Two blocks are connected by massless string that is wrapped around a pulley. Block 1 has a mass m1=4.20 kg, block 2 has a mass m2=1.30 kg, while the pulley has a mass of 2.10 kg and a radius of 27.4 cm. The pulley is frictionless, and the surface mass 1 is on is also frictionless.

If the blocks are released from rest, how far will block 2 fall in 3.00 s?

"Took the course in 1955" so you're allegedly in your 80's and have become so wise in the last 60 years that you now spend your time chiding teenagers on the internet? Nice.

boomer

I'm sorry but a question was asked because one couldn't figure it out. You can't say that they never tried when you weren't there with them to do the assignment. All they're asking for is a little help to see what they should do, or maybe if they got the question right. I too have a similar question like this and I did it but want to make sure it's right. If someone who actually wants to help, please tell us the steps of what to do. Thank you in advance.

dude u got problems

To find how far block 2 will fall in 3.00 s, we need to analyze the forces acting on the system.

First, let's understand the forces acting on each individual block.

For block 1:
- The force of gravity is acting downwards with a magnitude of m1 * g, where g is the acceleration due to gravity (approximated as 9.8 m/s^2).
- The tension in the string is acting upwards.

For block 2:
- The force of gravity is acting downwards with a magnitude of m2 * g.
- The tension in the string is acting upwards.

Now, let's analyze the forces acting on the pulley.

Since the pulley is frictionless, the only force acting on it is the tension in the string, which produces a torque (rotational force) on the pulley.

The torque caused by the tension in the string can be calculated as the tension multiplied by the radius of the pulley. Since the radius is given as 27.4 cm, we need to convert it to meters by dividing it by 100 (1 m = 100 cm).

Once we calculate the torque on the pulley, we can use Newton's second law for rotation (τ = I * α) to find the angular acceleration (α) of the pulley. Here, τ is the torque and I is the moment of inertia of the pulley.

The moment of inertia of a solid cylinder (which approximates the pulley) rotating about its central axis is given by I = (1/2) * m * r^2, where m is the mass of the pulley and r is its radius.

With the angular acceleration (α), we can find the linear acceleration (a) of block 2. Since the linear acceleration of block 1 and block 2 are the same (due to the massless and frictionless pulley), we can find the acceleration of both blocks.

Finally, we can use the kinematic equation for distance (s = ut + 0.5 * a * t^2) to find how far block 2 will fall in 3.00 s. Here, u is the initial velocity (which is 0, as the blocks are released from rest), a is the acceleration, and t is the time (3.00 s).

By following these steps, you can calculate how far block 2 will fall in 3.00 s.

I outlined one problem for you. However I see no sign that you have even made an attempt to do these. I took the course in 1955. I do not need the practice as badly as you do.