Solving the 8kg-5kg Pulley System: Calculating Acceleration & Time

[cowgiljl]

The question is an 8.00 kg block (A) resting on a horizonal surface is attached to a 5.00kg block (B) that hangs freely by a string over a pulley. The coefficient of friction for block(A) is .200
a) calculate the acceleration of the system?
b) how long does it take the blocks to move 20 meters?

I drew a picture adn this is what i have using the formula for accel.
A = M2g-Mk*M1g/M1+M2 which gave me 5 (9.80)-.200*(8)*(9.80)/ 5+8 = 2.56 m/s^2

that was all i was able to get was i right so far and could help me on finding where i went wrong on this problem

 

[jamesrc]

What you did is correct (though you should be more careful with your parenthesis for the sake of those of us reading your post). To find the answer to part b, use your equations for constant acceleration motion. I would assume this system starts at rest, so s - so = vo*t + .5*a*t^2 would be the way to go, with vo = 0.

 

[M98Ranger]

Great job on setting up the problem and using the correct formula for acceleration! Your answer of 2.56 m/s^2 looks correct. To double check, we can also use Newton's Second Law (F=ma) to solve for acceleration. In this case, the net force acting on the system is the weight of the hanging block (5kg * 9.8m/s^2) minus the friction force (0.2 * 8kg * 9.8m/s^2). This gives us a net force of 39.2 N. Plugging this into F=ma, we get a = 39.2 N / 13 kg = 3.01 m/s^2, which is very close to your answer of 2.56 m/s^2.

To find the time it takes for the blocks to move 20 meters, we can use the kinematic equation d=1/2at^2, where d is the distance, a is the acceleration, and t is the time. Plugging in the values we know, we get 20m = 1/2 * 2.56 m/s^2 * t^2. Solving for t, we get t = 3.16 seconds. So it would take approximately 3.16 seconds for the blocks to move 20 meters.

Great work on this problem! Keep practicing and you'll become a pro at solving pulley systems in no time.