Using a motion sensor to monitor a mass hanging on a spring one can produce a gr

Question

Using a motion sensor to monitor a mass hanging on a spring one can produce a graph of the vertical velocity of the mass as a function of time. The y-axis is shown in the figure with the positive direction indicated by the arrowhead near the ring stand. For this problem, ignore friction and air drag. 4s Answer the questions below by choosing all the correct answers and putting them in the boxes at the right of the questions. 3. At-0.4 s when the velocity is zero, the position of the mass is ZU·BAtitss low -TA BAt is low isz A. At its lowest point. B. At its highest point C. You can't tell from the information giver D Atequilibrium-going-up. E. At equilibrium going down 4. At-0.8 s when the velocity is at a positive maximum, the position of the mass is A. At its lowest point. B. At its highest point C. You can't tell from the information given. isatapos D At equilibrium going up. -E-At-equilibriumrgomgdomm - 5. At -0.8 s when the velocity is at a positive maximum, the elastic force of the spring on the mass is A. Pointing up (positive -B-Pointing-dewn (negative C. Zero D. You can't tell from the information given. TS At-0.15 s, when the velocity is at a negative maximum, the position of the mass is A. At its lowest point. 6. D. At equilibrium going up E. At equilibrium going down C. You can't tell from the information given. At-0.8 s when the velocity is at a positive maximum, the net force on the mass is B. Pointing down (negative) 7. C. Zero D. You can't tell from the information given. At-0.4 s when the velocity is zero, the force of the spring on the mass is A. Pointing up (positive) B. Pointing down (negative) 8. D. You can't tell from the information given.

Explanation / Answer

3-a   velocity is changing from - ve to positive so at lowest point

4-b  velocity from +ve to - ve so highest point

5-b acceleration=dv/dt =-ve at t=0.8s

6-e at t=0.15 velocity is at maximum positive according to given graph and dv/dt is changing from +ve to - ve so negative acceleration

7-b At t=0.8 velocity is zero according to given graph and a=dv/dt=-ve so force is downward

8-a a=dv/dt=+ve at t=0.4 s so force upward

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