# Physics - drwls can you help me with this?

A 2.00kg block hangs from a rubber cord, being supported so that the cord is not stretched.
The unstretched length of the cord is 0.500m and it's mass is 5.00g.The "spring constant" for the cord is 100N/m. Block is released and stops at lowest point.

a) determine the tension in cord when block is at lowest point.
I'm not sure but I do know that
Sum F= T-mg= 0 at lowest point.
but is the T= mg and that's it?

b) what is length of cord in stretched position?
I think but I'm not sure that I can find it by using

However I don't think I can use this since I don't have v or L.

c) find speed of transversal wave in cord if block is at lowest position.

what I thought transverse speed was was from the equation:
vy= -omega*A cos(kx-omega*t) but I don't think I can use k that was given since It's not the same and I don't think I have omega either so how would I find transversal wave speed?

Thanks alot

1. 👍 0
2. 👎 0
3. 👁 190
1. a) The block stops and turns around when it has traveled twice the equilibrium deflection.
When the block has reached its lowest elevation, lost gravitational potential energy of the block and cord is converted to potential energy of the stretched cord. Let X be the deflection from the starting position at that time.
(1/2) k X^2 = M g X + m g X/2
The second term is the loss in PE due to the lowering of the center of gravity of the cord. m is the cord's mass and M is the block's mass.
X = 2(M + m/2)g/k = 0.393 m
The cord tension is the spring constant times that deflection, or kX = 39.3 N
You cannot apply your T = Mg equation because the Mass is accelerating at the lowest position.
(b) Use the value of X I derived in (a).
(c) The speed of a transverse wave in a stretched cord is
V = sqrt [T/(m/L)], where
m/L is the cord mass per unit length. At lowest position, L = 0.500 + 0.393 = 0 893 m
m = 0.005 kg and T = 39.3 N
Solve for V

1. 👍 0
2. 👎 0
posted by drwls
2. Oh..I think one of the biggest problems with this question was my ability to actually visualize what was happening with the block. I thought they meant that it was at rest at it's lowest postion - thus it stopped moving and therefore SumF= 0.

Thanks very much for your help drwls =D

1. 👍 0
2. 👎 0
3. The block IS temporarily at rest at its lowest position, but it is accelerating up again at that time. It is not in force equilibrium there.

1. 👍 0
2. 👎 0
posted by drwls
4. So technically it would be this:

SumF= T-mg= ma right?

1. 👍 0
2. 👎 0
5. Yes. But you don't need to use that equation to get Tension or position vs. time. I used an energy method

1. 👍 0
2. 👎 0
posted by drwls
6. Yes, I saw that. But how do know to use energy equation and when do I use that one?

1. 👍 0
2. 👎 0
7. Using the energy conservation is always a good shortcut to get conditions at turnaround points, or maximum velocities, when there is oscillatory motion. You would used Newton's law of acceleration if you wanted to solve for the equation of motion vs. time. The quickest way to get the tension T in your problem is just to use T = kX and solve for the maximum X.

1. 👍 0
2. 👎 0
posted by drwls
8. Oh okay.

Thanks very much drwls =)

1. 👍 0
2. 👎 0

## Similar Questions

1. ### physics

5. A 2.00-kg block hangs from a rubber cord, being supported so that the cord is not stretched. The unstretched length of the cord is 0.500 m, and its mass is 5.00 g. The “spring constant” for the cord is 100 N/m. The block is

asked by arjunsareen on January 24, 2012
2. ### physics

A block of mass M hangs from a rubber cord. The block is supported so that the cord is not stretched. The unstretched length of the cord is L0 and its mass is m, much less than M. The "spring constant" for the cord is k. The block

asked by Stephen on December 1, 2009
3. ### Physics

A block of mass M hangs from a rubber cord. The block is supported so that the cord is not stretched. The unstretched length of the cord is L_0 and its mass is m, much less than M. The "spring constant" for the cord is k. The

asked by Keith on December 2, 2009
4. ### Physics

A stationary arrangement of two crayon boxes and three cords. Box A has a mass of 11.0kg and is on a ramp at angle è=30.0 degrees; box B has a mass of 7.00 kg and hangs on a cord. The cord connected to Box A is parallel to the

asked by Carla on December 8, 2006
5. ### Physics

A block of mass 5.53 kg lies on a frictionless horizontal surface. The block is connected by a cord passing a. over a pulley to another block of mass 2.25 kg which hangs in the air. Assume the cord to be light (massless and

asked by Emeera on February 21, 2014
6. ### physics

A block of mass 3.62 kg lies on a frictionless horizontal surface. The block is connected by a cord passing over a pulley to another block of mass 7.8 kg which hangs in the air, as shown. Assume the cord to be light (massless and

asked by alice on November 15, 2012
7. ### Physics

A small block with a mass of m = 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface. The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The

asked by Sonya on May 20, 2014
8. ### physics

A block (mass = 2.9 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1.2 x 10-3 kg·m2), as the figure shows. Initially the pulley is prevented from rotating and the block is stationary.

asked by danica on November 3, 2010
9. ### Physic

A block (mass 2.4 kg) is hanging from a massless cord that is wrapped around a pulley -3 2 (moment of inertia of the pulley = 1.3 x 10 kg·m ), as the drawing shows. Initially the pulley is prevented from rotating and the block is

asked by Jess on February 4, 2013
10. ### Physics-Mechanics

horizontal surface. The block is connected by a cord passing over a pulley to another block of mass 5.53 kg which hangs in the air, as shown on the following picture. Assume the cord to be light (massless and weightless) and

asked by Genevieve on September 22, 2010

More Similar Questions