A 7 kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The spring constant of the spring is 1.7 N/cm. The mass is pulled downward 2.8 cm and released. What is the speed of the mass when it is half of the distance above the point from which it was released? For example, if the mass was pulled till 2 cm, then half of the distance is 1 cm.Note: Write your answer in normal format up to 4 decimal points. Example: 12.3456. Please DO NOT write the units in the electronic format. Definitely, write the units on paper. If I don't find the units in your hand-written notes, then marks will be deducted.
Question
A 7 kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The spring constant of the spring is 1.7 N/cm. The mass is pulled downward 2.8 cm and released. What is the speed of the mass when it is half of the distance above the point from which it was released? For example, if the mass was pulled till 2 cm, then half of the distance is 1 cm.Note: Write your answer in normal format up to 4 decimal points. Example: 12.3456. Please DO NOT write the units in the electronic format. Definitely, write the units on paper. If I don't find the units in your hand-written notes, then marks will be deducted.
Solution
To solve this problem, we need to use the principle of conservation of energy. The total mechanical energy of the system (kinetic energy + potential energy) is conserved because there are no non-conservative forces doing work.
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First, we calculate the potential energy (PE) at the point from which the mass was released. This is given by the formula PE = 0.5 * k * x^2, where k is the spring constant and x is the displacement from the equilibrium position. Here, k = 1.7 N/cm = 170 N/m (since 1 m = 100 cm) and x = 2.8 cm = 0.028 m. So, PE = 0.5 * 170 * (0.028)^2 = 0.06724 J.
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Next, we calculate the potential energy at the point where the mass is half of the distance above the point from which it was released. Here, x = 0.028/2 = 0.014 m. So, PE = 0.5 * 170 * (0.014)^2 = 0.01681 J.
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The difference in potential energy is the kinetic energy (KE) at the point where the mass is half of the distance above the point from which it was released. So, KE = 0.06724 - 0.01681 = 0.05043 J.
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Finally, we calculate the speed of the mass using the formula KE = 0.5 * m * v^2, where m is the mass and v is the speed. Here, m = 7 kg. So, v = sqrt((2 * KE) / m) = sqrt((2 * 0.05043) / 7) = 0.3810 m/s.
So, the speed of the mass when it is half of the distance above the point from which it was released is approximately 0.3810 m/s.
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