Question

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 44.0° above the horizontal. The glider has mass 8.00×10^−2 kg . The spring has k = 580 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.60 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. (a) What distance was the spring originally compressed? (b) When the glider has traveled along the air track 0.60 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

Answer 1

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