Answer:
v = - 1,715 m / s
, x = 0.0156 m
Explanation:
This is an oscillatory movement exercise, which is described by the expression
x = A cos (wt + Ф)
we can assume that the block is released from its maximum elongation, so the phase constant (Ф) is zero
As we are told that the stone does not affect the movement of the spring mass system, the amplitude and angular velocity do not change, in the upward movement the stone is attached to the mass, but in the downward movement the mass has an acceleration greater than g leave the stone behind, let's look for time, for this we use the definition of speed and acceleration
v = dx / dt
v = - A w sin wt
a = - Aw² cos wt
a = -g
-g = - Aw² cos wt
wt = cos⁻¹ (g / Aw²)
t = 1 / w cos⁻¹ (g / Aw²)
angular velocity and frequency are related
w = 2π f
w = 2π 4
w = 8π rad / s
remember that the angles are in radians
t = 1 / 8π cos⁻¹ (9.8 / (0.07 64π²))
t = 0.039789 1.3473
t = 0.0536 s
let's find the speed for this time
v = - A w sin wt
v = - 0.07 8π sin (8π 0.0536)
v = - 1,715 m / s
the distance is
x = A cos wt
x = 0.07 cos (8π 0.0536)
x = 0.0156 m
