Newtons Laws - Boxes connected with a cord

  • Thread starter hsestudent
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  • #1

Homework Statement


A block A (mass 2.25k) rests on a tabletop. It is connected by a horizontal cord passomg over a light, frictionless pulley to a hanging block B (mass 1.3kg). The coeffisient of kinetic friction between block A and the tabletop is 0.45. After the block is released from the rest,
fint the speed of each block after moving 3cm.


Homework Equations





The Attempt at a Solution


I found the frictionforce by:

Frictionforce = coeffisient * N (where N = G of block A).

Frictionforce = 9,93 N.

The force on the rope will be = the G (gravityforce) on block B.

ForceOnRope = 12,75N

Then I tried to use Newtons second law to find the acceleration, but this attempt failed when I tried to find the velocity of the boxes, which will be the same?
 

Answers and Replies

  • #2
Doc Al
Mentor
45,179
1,500
The force on the rope will be = the G (gravityforce) on block B.

ForceOnRope = 12,75N
The tension in the rope does not equal the weight of block B. If it did, then the blocks would be in equilibrium.

Call the tension T. Set up equations for both blocks and you'll be able to solve for the tension and the acceleration.
 
  • #3
39
0
tension

Note that the Tension will be the same throughout the rope. So if you set up the forces on each block and solve them for the Tension, you can then set those 2 equations equal to solve for other variables, such as the acceleration. Then you can use kinematics to solve for the speed.
 
Last edited:
  • #4
Thanks, will try this tomorrow morning!
 

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