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 (Figure 1). What is her mass if the spring constant is 280 N/m ? Express your answer to two significant figures and include the appropriate units. What is her speed when the spring's length is 1.1 m ? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

a)As observed from the figure, the period of oscillation is
3.0 s and as we know that

T = 2pi m/k

m = T^2K/4pi^2 = 9*280/4*3.141^2 = 63.85Kg

(b) What is her speed when the spring's length is 1.06 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 = 1/2 KA^2 = 0.5*280*0.4^2 = 22.4J

When the length of the spring is 1.1 m its elongation is 1.1 -1.0 = 0.1 m, hence the
elastic potential energy in this position is given by
UT = 1/2 KA^2 = 0.5*280*0.1^2 = 1.4J

Thus according to law of conservation of energy
Gain in kinetic energy = loss in Potential energy
Or KE at this position =UT-U = 22.4-1.4 = 21J

This gives the speed v at this position as

1/2 mv^2 = KE

v = 2*KE / m = 2*21/63.85 = 0.811m/s

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