How Long Does It Take a Block to Move 30cm with a Frictionless Pulley System?

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Homework Help Overview

The problem involves a block connected to a frictionless pulley system, where a constant tension is applied. The objective is to determine the time it takes for the block to move a specified distance of 30 cm, starting from rest.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply Newton's second law and kinematic equations to find the time taken for the block's movement. Some participants suggest considering the torque on the pulley and its relationship to angular acceleration. Others explore the connection between linear and angular motion.

Discussion Status

Participants are actively discussing the role of torque in the problem and how it relates to the block's motion. There is a recognition of the methods used, but no explicit consensus on the correctness of the calculations has been reached.

Contextual Notes

There is a mention of the original poster's uncertainty regarding the application of torque and its effects on the system. The discussion reflects a mix of interpretations and approaches to the problem without resolving the final answer.

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A 1.5 kg block is connected by a rope across a 50 cm-diameter, 2.0 kg, frictionless pulley. A constant 10 N tension is applied to the other end of the rope. Starting from rest, how long does it take the block to move 30cm?

I summed up the force of the tension and the force of gravity of the block. I defined the direction the tension was going to be positive which would make the force of gravity be negative. Summing them up I got -4.7 N and then using Newton's 2nd law I found the acceleration to be -3.13 m/s^2. Knowing Vo=0, d=.3m, a=-3.13 m/s^s and using the kinematic equation: d=Vo*t+.5*a*t^2. Plugging in everything I know I got the time to be .44s. This wasn't the right answer and I have no idea what to do now. I'm thinking that the torque on the pulley might be involved but I can't tell how to use it.
 
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Yup you got it right...the torque will be involved.
We know that the torque is the product of force and perpendicular distance. Hence, the torque on the pulley will be the tension applied x the radius of the pulley (in m). Draw a FBD again taking this into account. Let us know if you get stuck.
 
I found the torque to be 2.5 N*m. Then using T=I\alpha I found the angular acceleration to be 20 rad/s^2. Then using s=\Theta*r I found the distance the pulley moved was 1.2 radians. I then used the circular motion for of the kinematic equation I used before and got .35s. Does this seem right?
 
That does seem right...I am not talking about the calculations, but the method you employed.
Though I think that it could be solved by taking linear acceleration from the angular acceleration too.
Anyways, have you checked your answer?
 

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