How Is Velocity Calculated for a 5kg Block in a Pulley System with a Spring?

[zd1899]

1. A pulley has a massless in extensible thread whose one side has a 5 kg block along with a spring of k=40 with it and the other end has a 2kg block (which is on the ground ). Find velocity of the 5kg block when the 2 kg block moves up.

The Equations part :
taking Tension as T , and extension cause by 5kg block as x1 and 2kg block as x2
we have:
T-kx1-50=Kx2-T+20


That includes my attempt as well.
I could find out the x1+x2 = 7/4 but then the velocity with that comes around as 3.45 m/s , but the correct answer is 1.414 m/s <2^1/2>.

 

[Hootenanny]

You seem two have omitted sections two and three from the homework template.

 

[nlightner]

Hello,

Thank you for reaching out for help with your problem involving motion with pulleys. I am happy to assist you in finding the correct solution.

First, let's analyze the situation and understand what is happening. We have a pulley with a massless, inextensible thread that is connected to a 5kg block and a spring with a spring constant of 40. On the other end of the thread, there is a 2kg block on the ground. We are trying to find the velocity of the 5kg block when the 2kg block moves up.

To solve this problem, we can use the equations of motion for a system of connected masses. We can start by considering the forces acting on the 5kg block. There is the force of gravity pulling it down, and the tension in the string pulling it up. We can also consider the force from the spring, which is given by Hooke's Law as F = kx, where k is the spring constant and x is the extension of the spring.

Next, we can consider the forces acting on the 2kg block. There is only the force of gravity pulling it down, as the tension in the string is acting in the opposite direction.

Now, let's set up our equations. We can use the fact that the tension in the string is the same throughout the string, and we can also use the fact that the extension of the spring is the same on both sides of the pulley. This gives us the following equations:

For the 5kg block:
T - mg - kx = ma

For the 2kg block:
mg = ma

Next, we can substitute the value of the acceleration, a, from the second equation into the first equation. This gives us:

T - mg - kx = mg

Rearranging this equation, we get:

T = 2mg + kx

Now, we can use the fact that the extension of the spring is the same on both sides of the pulley. This means that x = x1 + x2, where x1 is the extension caused by the 5kg block and x2 is the extension caused by the 2kg block. Substituting this into our equation for T, we get:

T = 2mg + k(x1 + x2)

We also know that the total mass of the system is 7kg (5kg + 2kg).