Question

A mass of 0.500 kg stretches a spring 20.0 cm. Find the force constant of the spring. If the mass is pulled down an additional distance of 5.00 cm and then released, find the initial restoring force, the period, and the total energy of the resulting simple harmonic motion.

2- Elevator problem with scale (no mass of elevator given). Find what the scale reads if (a) it goes up at 2m/s, (b) up at 8m/s^2 (c) down at 5m/s^2.

PLEASE ANSWER BOTH QUESTIONS AS I DON'T HAVE ANY REMAINING QUESTIONS FOR THIS MONTH.

THANK YOU!

Answer 1

1. The force on the string is the gravitational force, Therefore by Hooke's Law we can write,

The restoring force when it is stretched by 5 cm is,

The total energy will be the initial potential energy of the spring which is,

2. If it goes at a constant velocity in the upward direction, then the scale will be showing the same value as when it is at rest. Let the mass of person standing be m, then the reading of the scale will be mg.

When the elevator is going up by 8m/s^2 then the reading will be m(g + 8) = 17.81m

When the elevator is going down by 5 m/s^2 then the reading will be m(g-5) = 4.81m