Help ly need someone to help me check pulley problem

Click For Summary
SUMMARY

The discussion focuses on a physics problem involving two masses connected by a pulley system, specifically a 1.00 kg mass on the left and a 6.50 kg mass on the right, which is 2.00 m above the ground. The user attempts to calculate the speed of each mass just before the heavier mass strikes the ground using energy conservation principles. The correct equation involves ensuring proper symbol usage for mass and height, and the user is confirmed that their approach is valid despite initial symbol errors.

PREREQUISITES
  • Understanding of Newtonian mechanics
  • Familiarity with energy conservation principles
  • Basic algebra for solving equations
  • Knowledge of pulley systems and their dynamics
NEXT STEPS
  • Study energy conservation in mechanical systems
  • Learn about pulley mechanics and tension forces
  • Explore kinematic equations for motion analysis
  • Practice solving problems involving multiple masses and pulleys
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in solving problems related to pulleys and energy conservation.

squintyeyes
Messages
44
Reaction score
0
Help urgently need someone to help me check pulley problem

Two masses are connected over a small pulley using a light string. The mass on the left is 1.00 kg and the mass on the right is 6.50 kg. The mass on the right is a distance 2.00 m above the ground. The system is released from rest as shown in the animation below.



What is the speed of each object just before the mass on the right strikes the ground?
________________m/s

Attempt
mgh + (mv^2)/2 + (Mv^2)/2 = mgh + (mV^2)/2 + (Mv^2)/2
(6.5)(9.8)(2) + 0 + 0 = 1(9.8)(2) + (6.5v^2)/2 + V^2/2
(6.5)(9.8)(2) = 1(9.8)(2) + (7.5v^2)/2

If i solve for this will i get the right answer?
 
Physics news on Phys.org


squintyeyes said:
Attempt
mgh + (mv^2)/2 + (Mv^2)/2 = mgh + (mV^2)/2 + (Mv^2)/2
(6.5)(9.8)(2) + 0 + 0 = 1(9.8)(2) + (6.5v^2)/2 + V^2/2
(6.5)(9.8)(2) = 1(9.8)(2) + (7.5v^2)/2

If i solve for this will i get the right answer?

Yes. You switched some symbols in the first line which should be:

mgh + (mv^2)/2 + (Mv^2)/2 = Mgh + (mV^2)/2 + (MV^2)/2

but you substituted all the right constants anyway.
 


thanks
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
30
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K