Ask Question
29 April, 00:41

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?

+4
Answers (1)
  1. 29 April, 01:22
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers