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A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.76 m and whose unstretched length is 0.41 m. Next the mass is pulled down to where the spring has a length of 1.01 m and given an initial speed upwards of 1.6 m/s. What is the maximum length of the spring during the motion that follows?

Respuesta :

Answer;

= 1.256 m

Explanation;

We can start by finding the spring constant  

F = k*y  

Therefore;  k = F/y = m*g/y

                               = 0.40kg*9.8m/s^2/(0.76 - 0.41)

                               = 11.2 N/m  

Energy is conserved  

Let A be the maximum displacement  

Therefore;  1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2  

Thus;  A = sqrt((1.20 - 0.55)^2 + m/k*v^2)

               = sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)

                = 0.846 m  

Thus; the length will be 0.41 + 0.846  = 1.256 m

onyebuchinnaji

The maximum length of the spring is 1.31 m.

Spring constant

The spring constant of the spring is calculated as follows;

[tex]F = kx\\\\k = \frac{F}{x} \\\\k = \frac{mg}{x} \\\\k = \frac{0.4 \times 9.8}{0.76 - 0.41} \\\\k = 11.2 \ N/m[/tex]

Displacement of the spring

The displacement of the spring during the upward motion is calculated as follows;

[tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2\\\\x^2 = \frac{mv^2}{k} \\\\x = \sqrt{\frac{mv^2}{k} } \\\\x = \sqrt{\frac{0.4 \times 1.6^2}{11.2} }\\\\x = 0.3 \ m[/tex]

Thus, the maximum length of the spring is (0.3 + 1.01) = 1.31 m.

Learn more about spring constant here: https://brainly.com/question/1968517

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