If a mass of 48 kg is hung on a spring, k = 2300 N/m, what would the period of o

Question

If a mass of 48 kg is hung on a spring, k = 2300 N/m, what would the period of oscillation be? In the lab Springs, a student sits on a bosun's chair suspended by a spring. If the has a spring constant, k = 1100 N/m, and is stretched by 0.5 m, how-much energy is stored in the spring? The student gets off the bosun's chair very carelessly and all the energy in the spring is converted into the kinetic energy of the seat. The mass of the seat is 1 kg. Once the spring has fully contracted, what is the speed of the seat? Unfortunately, the student was hit on the head by the seat as it flew up. Some of kinetic energy of the seat will be converted into gravitational potential energy; neglect this. Assume the impact speed of the seat is the same as in the previous question. If the seat comes to rest in 0.01 it hits the student's head, what force would the seat exert on the student's head? Be safe. Remember: get off the bosun's chair carefully!

Explanation / Answer

5. Use equation,

T = 2*sqrt(m/k) = 2*3.14*sqrt(48/2300) = 0.91s

6. Use equation

SPE = 1/2kx^2 = ½*1100*0.5^2 = 137.5 J

7. Use law of conservation of energy

PE=KE

1/2kx^2 = 1/2mv^2

v= x*sqrt(k/m)

k= 1100N , x= 0.5m

v= x*sqrt(k/m) = 0.5*sqrt(1100/1) = 16.6 m/s

8. F = J/t = [m(vf-vi)]/t = [1(0-16.6)]/0.01 = -1660 N

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