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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 released and stops momentarily at the lowest point.

(a) Determine the tension in the cord when the block is at this lowest point.
(b) What is the length of the cord in this "stretched" position?
(c) If the block is held in this lowest position, find the speed of a transverse wave in the cord.

Respuesta :

Answer:

a. T = 19.6N  

b. L2 = 0.696m  

c. v= 75.04m/s

Explanation:

Mass of block = m1 = 2.0kg

unstretched length of the cord= L1 =0.50m

Mass of the cord= m2 = 5.00g = 5.00/1000 = 0.005kg

Spring constant for the cord = k = 100N/m

Acceleration due to gravity = g = 9.8m/sec2

Part a.Tension T in the cord when the block is in the lowest point

as tension in the cord and the weight of the block are acting in opposite direction

for the block to be in equilibrium position sum of forces acting on it must be zero

∑〖F=0〗  

T+(-W)=0 (because W is opposite in direction to T)

T=W  

T=mg (m shows the mass of the block)

Putting numerical values, we get

T=2×9.8

T=19.6N  

Part b. The length of the cord in the "stretched" position

let x be the stretched length of the rubber cord

then according to Hook’s law,  

T=kx (T =tension in the rubber cord)

Re-arranging the equation, we get

x=T/k

Putting numerical values, we get

x=19.6/100  

x=0.196m  

Length of the stretched cord =L2=L1+x

Length of the stretched cord=L2=0.50+0.196=0.696m  

Part c. If the block is held in this lowest position, find the speed of a transverse wave in the cord.

As we know that the formula for the speed of a wave on a string is the square root of the tension in the string per unit mass of the string per unit length of the string i.e.

v=√((Tension in the string )/((mass of the string)/(length of the string)))  

v=√(T/(m2/L2))  

Putting numerical values, we get

v=√(19.6/(0.005×0.696))  

v=√(19.6/0.00348)  

v=√5632.18  

v=75.04m/s  

 

 

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