contestada

You and a friend are watching people on a new carnival ride. The ride begins with a car and rider ( 150kg combined) at the top of a curved track. At the bottom of the track is a 50-kg block of cushioned material that is attached to a horizontal spring whose other end is fixed in concrete. The car slides down the track ending up moving horizontally when it crashes into the cushioned block, sticks to it, and oscillates at a rate of 3 repetitions in 10s.
Just after the collision, you notice that the spring's maximum compression is 2 meters. From what height did the car start?

Respuesta :

jacob193

Answer:

Approximately [tex]0.97\; {\rm m}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)

Explanation:

Overview of steps:

  • Using info about the oscillation, find the spring constant.
  • Apply conservation of energy to find height change.

After sticking to the cushioned block on the spring, the combination is a system in a simple harmonic motion:

  • Mass of oscillator: [tex]50\; {\rm kg} + 150\; {\rm kg} = 200\; {\rm kg}[/tex] (assuming mass of spring is negligible.)
  • Period of oscillation: [tex]T = (10\; {\rm s}) / 3 = (10/3)\; {\rm s}[/tex].

Let [tex]A[/tex] denote the amplitude of the motion. At time [tex]t[/tex]:

  • Position would be [tex]x(t) = A\, \sin((2\, \pi / T)\, t)[/tex].
  • Velocity would be [tex]\omega(t) = A\, (2\, \pi / T)\, \cos((2\, \pi / T)\, t)[/tex]..
  • Acceleration would be [tex]a(t) = - A\, (2\, \pi / T)^{2}\, \sin((2\, \pi / T)\, t)[/tex]
  • Restoring force would be [tex]F(t) = m\, (a(t)) = - m\, A\, (2\, \pi / T)^{2} \, \sin((2\, \pi / T)\, t)[/tex].

Since by Hooke's Law [tex]F = -k / x[/tex] for this spring, the spring constant [tex]k[/tex] would be:

[tex]\begin{aligned}k &= - \frac{F(t)}{x(t)} \\ &= m\, \left(\frac{2\, \pi}{T}\right)^{2} \\ &= (200\; {\rm kg})\, \left(\frac{2\,\pi}{(10/3)\; {\rm s}}\right)^{2} \\ &\approx 710.612\; {\rm N\cdot m^{-1}}\end{aligned}[/tex].

Let [tex]x = 2\; {\rm m}[/tex] denote maximum displacement. Elastic potential energy:

[tex]\begin{aligned} (\text{EPE}) &= \frac{1}{2}\, k\, x^{2} \\ &\approx \frac{1}{2}\, (710.612)\, (2)^{2}\; {\rm J} \\ &\approx 1.42122\times 10^{3}\; {\rm J}\end{aligned}[/tex].

Let [tex]\Delta h[/tex] denote the change in height. The change in gravitational potential energy would be [tex](\text{GPE}) = m\, g\, \Delta h[/tex] where [tex]m = 150\; {\rm kg}[/tex].

Assuming that energy is conserved, and [tex](\text{EPE})[/tex] is equal to the change in [tex](\text{GPE})[/tex]. The change in height would be:

[tex]\displaystyle \frac{1.42122\times 10^{3}\; {\rm J}}{(150\; {\rm kg})\, (9.81\; {\rm N\cdot kg^{-1}})} \approx 0.97\; {\rm m}[/tex].