Question

A block with mass m =7.3 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.29 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.9 m/s. The block oscillates on the spring without friction.

1) What is the spring constant of the spring?

2) What is the oscillation frequency?

3) After t = 0.45 s what is the speed of the block?

4) What is the magnitude of the maximum acceleration of the block?

5) At t = 0.45 s what is the magnitude of the net force on the block?

6) Where is the potential energy of the system the greatest? (Choose all that apply)

[ ] At the highest point of the oscillation.

[ ] At the new equilibrium position of the oscillation.

[ ] At the lowest point of the oscillation.

PLEASE SHOW STEPS

Answer 1

Mass of the block is, m 7.3kg Distance that the spring stretches is, x = 0.29 m Initial velocity of the of the block is, v 4.9m/s Solution: Expression for the spring force is, F-mg kx=mg Rearrange the above equation to get k, 7.3x9.8 0.29 246.7 Nm The spring constant of the spring is 246.7 N/m

Expression for the oscillation frequency is, 1 246.7 0.925Hz The oscillation frequency is 0.925Hz Positon of SHM of a particle is, x-Asin ot Velocity of SHM of a particle is, doc dt rom the work energy theorem.

Rearrange the above equation to get X nv 7.3x4.92 246.7 0.843m Amplitude of the wave is A-0.843m Angular velocity of the wave is, -2π(0.925) 5.81rad/s Velocity of the SHM is, --(0.843)x(5.81)x cos (5.81t) t 0.45s v-(-4.9)(-0.864) - 4.23m/s The speed of the block is v- 4.34m/s

Maximum acceleration of the SHMis -(5.81) (0.843) -28.5m/s Maximum acceleration of the SHM is 28.5m/s Net force on the block is, 045 ma magu sin(t) 7.3x28.5x sin (5.81x0.45) 104.53N Net force on the block is 104.53N Potential energy depends on the extension and the compression of the spring At highest and lowest points, the extension and compression is maximum. Hence, the potential energy of the system is highest.