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
v= rad(T/mu)= rad(mgL/mblock)
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
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
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.
So technically it would be this:
SumF= T-mg= ma right?
Yes. But you don't need to use that equation to get Tension or position vs. time. I used an energy method
Yes, I saw that. But how do know to use energy equation and when do I use that one?
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.
Oh okay.
Thanks very much drwls =)
a) To determine the tension in the cord when the block is at its lowest point, you need to consider the forces acting on the block. At the lowest point, the tension in the cord will be equal to the weight of the block.
Thus, T = mg, where T is the tension in the cord, m is the mass of the block, and g is the acceleration due to gravity.
In this case, the mass of the block is given as 2.00 kg. Assuming standard gravity of 9.8 m/s^2, you can calculate the tension as follows:
T = (2.00 kg) * (9.8 m/s^2) = 19.6 N
Therefore, the tension in the cord when the block is at its lowest point is 19.6 N.
b) To find the length of the cord in the stretched position, you can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
The equation for Hooke's Law is given by F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the spring constant (k) for the cord is given as 100 N/m. The displacement (x) from the equilibrium position is equal to the unstretched length of the cord.
Therefore, the length of the cord in the stretched position is 0.500 m.
c) To find the speed of the transverse wave in the cord when the block is at its lowest position, you can use the formula for wave speed.
Wave speed (v) is given by the equation v = sqrt(T/μ), where T is the tension in the cord and μ is the linear mass density of the cord.
The linear mass density (μ) of the cord can be calculated using the mass of the cord (5.00 g) and its length (0.500 m).
μ = (mass of the cord)/(length of the cord)
= (5.00 g)/(0.500 m)
= 0.0100 kg/m
Substituting the values into the formula for wave speed:
v = sqrt(19.6 N / 0.0100 kg/m)
= sqrt(1960 m^2/s^2 / 0.0100 kg/m)
= sqrt(196000) m/s
= 140 m/s
Therefore, the speed of the transverse wave in the cord when the block is at its lowest position is 140 m/s.
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