Question

A block with mass m =6.7 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.6 m/s.

The block oscillates on the spring without friction.1) What is the spring constant of the spring? ANSWER = (234.5)2) What is the oscillation frequency? ANSWER = (0.942)3) After t = 0.47 s what is the speed of the block?4) What is the magnitude of the maximum acceleration of the block?5)At t = 0.47 s what is the magnitude of the net force on the block?6)Where is the potential energy of the system the greatest?---At the highest point of the oscillation.---At the new equilibrium position of the oscillation.---At the lowest point of the oscillation

Answer 1

The spring constant of the spring is approximately 234.50 N/m.

Step-by-step explanation:

To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. In this case, we know that the mass of the block is 6.7 kg and the displacement of the spring is 0.28 m. Therefore, we can use the equation F = kx, where F is the force, k is the spring constant, and x is the displacement. Plugging in the given values, we get:

F = kx
F = (6.7 kg)(9.8 m/s^2)
F = 65.66 N
65.66 N = k(0.28 m)
k = (65.66 N)/(0.28 m)
k ≈ 234.50 N/m

Therefore, the spring constant of the spring is approximately 234.50 N/m.