Maximum Height of Lighter Object in Pulley System - Homework Solution

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SUMMARY

The discussion focuses on calculating the maximum height reached by a lighter object in a pulley system involving two masses: 3.2 kg and 2.2 kg. The initial height of both masses is 1.80 m, and the pulley is positioned 4.8 m above the ground. The acceleration was determined to be 1.8 m/s², and the velocity of the lighter object was calculated to be 2.56 m/s just before it reaches its peak height. The final maximum height calculated was approximately 3.93 m, indicating that the lighter object does not hit the pulley during its ascent.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Basic kinematics equations, specifically y=yo+vo+1/2at²
  • Concept of gravitational acceleration (approximately 9.8 m/s²)
  • Knowledge of motion under constant acceleration
NEXT STEPS
  • Review kinematic equations for motion in one dimension
  • Study the principles of pulley systems and their dynamics
  • Explore the effects of mass and acceleration on object motion
  • Learn about energy conservation in mechanical systems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of pulley systems and the application of kinematic equations in real-world scenarios.

joseg707
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Homework Statement


Two masses, 3.2kg and 2.2kg are each initially 1.80m above the ground, and the massless and frictionless pulley is 4.8m above the ground. What maximum height dies the lighter object reach after the system is released? Assume it doesn't hit the pulley.


Homework Equations


F=ma
y=yo+vo+1/2at2

The Attempt at a Solution


I solved for acceleration by using F=ma and got 1.8m/s2.
I then solved for the velocity of the lighter object when the heavier object hit the ground by solving for the time it took for the heavier object to hit the ground and used that time to find the velocity

-1.8=-0.9t2
t=1.4

v=1.8*1.4
v=2.56

After finding the velocity I used that to find the maximum height.

y=2.56t-4.9t2
t=0, 0.522

0.522/2
t=0.261

y=3.6+2.56t-4.9t2
y=3.93333

I think I didn't take something into account or I used the wrong equation completely. Can someone help me?
 
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