Question

Part A 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 310 N/m? Express your answer to two significant figures and include the appropriate units. m = 71 kg Submit Previous Answers ✓ Correct Figure < 1 of 1 > Part B L (m) 1.4 1.2 What is her speed when the spring's length is 0.70 m ? Express your answer to two significant figures and include the appropriate units. 1.0 0.8 Åmoa ? 0.6 0.4 0.2- 0.0 + v= 0.6

Answer 1

a) k = 310 N/m from T time for one oscillatians 3. 0sec & we know that I = 20 m > m= KT² 402 - 3100 3² 452 b) from fig. Amplitude = maximum elongation A = 0.4m & it is also clear that natural length - oh spring is im - So maximum elastic P.E. or total enessy is Us įkA? = { x 310 * (04)? = 24.8 J - I when the length of spring is 0.70 m so the compression is 0.30 m Elastic potential Enessy in This position - Ust Kx2lgx 310 x 0.302= 13.95 J - By conservation of in KE = loss in Engsy P.E. Gain 1., K. Eat this post = 24-8-13-95 10.85 J _ mu= KE____ 2 ke Vm 2x10.85 V 71 * v= 0.55