Ablock with mass m =6.9 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 = 4.1 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 = 0.42 s what is the speed of the block?

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

5)at t = 0.42 s what is the magnitude of the net force on the block?

Answer mechanical energy;

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.