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

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A 1.5 kg block connected to a frictionless pulley system is subjected to a constant tension of 10 N, and the discussion revolves around calculating the time it takes for the block to move 30 cm. Initial calculations yielded an acceleration of -3.13 m/s², leading to an incorrect time of 0.44 seconds. The involvement of torque on the pulley was identified as crucial, with the torque calculated to be 2.5 N*m, resulting in an angular acceleration of 20 rad/s². After further calculations, a new time of 0.35 seconds was proposed, with the method used being validated as correct. The conversation emphasizes the importance of considering both linear and angular dynamics in solving the problem.
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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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