Answer:
a) x = 0.0175 m and b) v = 1.178 m/s
Explanation:
a) Let's analyze the exercise gives us a compressed spring, which would have elastic energy and a glider that is initially at rest and then released acquiring kinetic and potential energy, as there is no friction, the energy is conserved
Initial
Emo = Ue = ½ k x²
Final
Emf = K + U = ½ m v² + mg y
Emo = Emf
½ k x² = ½ m v² + mg y
At the highest point the speed is zero (v = 0). Let's calculate height trigonometry
sin 44 = y / L
y = L sin44
½ k x² = 0 + m g y
x² = 2m g L sin44 / k
x = √ (2 8.00 10⁻² 1.6 sin44 / 580)
x = 0.0175 m
b) When the glider is 0.60m it is no longer in contact with the spring that contracted only 0.0175m
Let's write the mechanical energy at this point
Em2 = K + U
Em2 = ½ m v² + mg y₂
L2 = 0.60 m
y₂ = L2 sin44
Emo = Em2
½ k x² = ½ mv² + mgh
v² = (½ k x² -mgh) 2 / m
v = √ [(½ k x² - mg L₂ sin44) 2 / m]
Let's calculate
v = √ [(½ 580 0.0175² - 8 10⁻² 0.6 sin44) 2/8 10⁻²]
v = √[ (0.08881 – 0.0333) 25]
v = 1.178 m/s