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In a local park, a pine cone falls off a tree branch (starting at rest) and falls to the ground. We will model the pine cone hitting the ground as if it is a mass going onto a spring (with the ground being a spring). The pine cone has a mass of 0.116 kg, and starts at a height of 7.90 m above the ground. The pine cone compresses the spring (ground) by 0.0148 m, briefly coming to rest at the bottom of its motion. (It then bounces, which is not part of the time period that we'll cover in this problem.) Assume that the pine cone, the Earth and the spring (ground) are a system, and that no net work is done by external forces to the system or by non-conservative forces. Note: these are the assumptions necessary to assume that mechanical energy is conserved. Label the initial time point as the start of the pine cone's drop, and the final time point as the moment when the pine cone is at rest with the spring as compressed as it will get.
1. What is the change in gravitational potential energy from the initial time point to the final time point? Hint: don't forget that the pine cone actually goes slightly below the height of most of the ground. Joules
2. What is the spring constant of the spring? N/m
3. Now, let's look at the time point when the pine cone is 4.24 meters above the ground. You can do this by assuming that this new time point becomes the final time point, and the spring is no longer in the problem. What is the magnitude of the pine cone's velocity at this time point? m/s

Sagot :

Answer:

1)) ΔU = -8.96 J, 2)    k = 8.18 10⁴ N / m, 3)  v = 8.47 m / s

Explanation:

For this exercise we will use conservation of energy.

Starting point. Point where the pineapple comes out

          Em₀ = U = m g h

where the reference frame is placed on the ground

Final point. Point where pineapple stops

          Em_f = K_e + U = ½ k y² + m g y

1) the change in gravitational potential energy is

           ΔU = U_f - U₀

           ΔU = m g y - m g h

           ΔU = mg (y-h)

         

let's calculate

            ΔU = 0.116 9.8 (0.0148 - 7.9)

            ΔU = -8.96 J

The negative sign indicates that the energy decreases

2) let's use energy conservation

             Em₀ = Em_f

             mg h = ½ k y² + mg y

             k = mg (h-y) [tex]\frac{2}{y^2}[/tex]

             

let's calculate

             k = 0.116 9.8 (7.9 - 0.0148)    [tex]\frac{2}{0.0148^2}[/tex]  

             k = 8.18 10⁴ N / m

3) we use the same starting point and as the end point we use this height (y₂ = 4.24 m)

             Em_{f2} = K + U = ½ m v² + mg y₂

             

energy is conserved

             Em₀ = Em_{f2}

              mgh = ½ m v² + m g y₂

              v =[tex]\sqrt{ 2g(h-y_2)}[/tex]

let's calculate

              v = [tex]\sqrt{ 2 \ 9.8 \ (7.9-4.24)}[/tex]

              v = 8.47 m / s

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