Question: Helppppp A mass hanging from a vertical spring is somewhat more complicated than a mass attached to a horizontal spring because the gravitational force acts

Helppppp

Helppppp A mass hanging from a vertical spring is somewhat more complicated

A mass hanging from a vertical spring is somewhat more complicated than a mass attached to a horizontal spring because the gravitational force acts along the direction of motion. Therefore, the restoring force of the oscillations is not provided by the spring force alone, but by the net force resulting from both the spring force and the gravitational force. Ultimately, however, the physical quantities of motion (position, velocity, and acceleration) for a vertical mass on a spring exhibit the same oscillations as a horizontal mass on a spring. A 100 g mass hangs from a vertical spring as shown in the picture. The measuring stick shows us the vertical y position of the bottom of the spring with the origin (y = 0) at the top of the spring. Note that the positive y direction is downward. The 100 g mass is at rest at the position shown (y = 50 cm). The dashed line in the picture (y = 30 cm) indicates the unstretched resting length of the spring. The mass is pulled down 6 cm, stretching the bottom of the spring to y = 56 cm, and then released so that it begins oscillating. 1. What are the equilibrium position and amplitude of the oscillations? y = L cm cmn 2. Is the spring force equal to zero at the equilibrium point? Explain why or why not. This answer has not been graded yet. 3. What is the spring constant? k = N/cm 4. What is the force of the spring on the mass (magnitude and direction) at both the highest point of the oscillation (smallest y) and the lowest point of the oscillation (largest y)? If the force is zero, enter 0 for the magnitude and choose No direction. spring (highest) = N, -Select- pspring (lowest) = N, -Select- 5. Draw a free body diagram for the mass when it's at the highest point of its oscillation. What direction is the net force on the mass at this point? If there is no net force, select No direction. -Select- Use your free body diagram and Newton's second law to find the magnitude of the accel ences at the highest point of its oscillation (this is the maximum acceleration). Iamaxl = 6. Below are shown a series of graphs associated with the motion of the mass and a series of physical quantities. The horizontal axis for every graph is time. The graph labeled (A) is a graph of the mass's position y (keep In mind that the positive direction is down, so the maximum y value corresponds to the lowest point of the oscillation). For each physical quantity Identify which graph could represent that quantity for this situation (if appropriate numerical scales were used on each axis). If none are possible, answer None. You may use a graph more than once. (Note that t = 0 on the graphs corresponds to some in-between point of the oscillation rather than a maximum, minimum, or zero. This means that we began recording the data on the graphs after the mass had already started oscillating. There would be some phase constant po, but this is not an important detail.) a. Velocity of the mass. -Select- b. Acceleration of -Select- V

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