Astronauts in space cannot weigh themselves by standing on a bathroom scale. Ins

Question

Astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating on a large spring. Suppose an astronaut attaches one end of a large spring to her belt and the other end to a hook on the wall of the space capsule. A fellow astronaut then pulls her away from the wall and releases her. The spring's length as a function of time is shown in the figure below. (a) What is her mass if the spring constant is 240 N/m? 54.7 kg (b) What is her speed when the spring's length is 1.28 m? 0.783 m/s

Explanation / Answer

(a) from figure time period T = 3 s

k = 240 N/m

T = 2pi(m/k)^0.5

3 = 2*3.14*(m/240)^0.5

m = 54.7 kg

(b) L =1.28 m

From the figure it is clear that the natural length of the spring is 1.0 m and the maximum elongation (Amplitude) is 0.4 m. Hence the maximum elastic potential energy or the total energy of the system is given by

UT = 0.5kA^2 = 0.5*240*0.4^2

UT = 19.2 J

When the length of the spring is 1.28 m its elongation is 1.28 -1.0 = 0.28 m,

hence the elastic potential energy in this position is given by

U = 0.5kx^2 = 0.5*240*0.28^2 = 9.408 J

From conservation of energy

gain in kinetic energy = loss in potential energy

KE at this position = UT - U

= 19.2 - 9.408 = 9.792 J

KE = 0.5mv^2

0.5*54.7*v^2 = 9.792

v = 0.598 m/s

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