Students have a small toy whose spring can be compressed. The students compress the spring a distance x from equilibrium (equilibrium length = Lo= 0.4 m). When they release the toy, it jumps up in the air. Consider air resistance to be negligible. a. Rank the total energy of the system from largest (1) to smallest (3) for when the toy is compressed and starts on the ground, when the toy is half-way to its maximum height h, and when the toy reaches the maximum height h. You do not need to use all of the numbers if the energy is the same at two or more of the heights. (e. g., 1, 1, 2) On the ground Half-way to height h At height h b. Describe how you used conservation of energy in making your choices. Students perform an experiment with the toy to find its spring constant k. They compress the spring multiple times to the same compression x and then release the spring, measuring the height h it reaches. They affix additional mass to the toy with each trial and record the combined mass m of the toy plus the extra mass. You may assume that all of the mass is in the top portion attached to the spring. With the spring compressed a distance x=0.020 m in each trial, the students obtained the following data for different values of m. m (kg) hm) 0.030 0.333 0.045 0.222 0.060 0.167 0.075 0.133 0.090 0.111

a) the total energy is always the same,  b )  k = 509.6 N / m

Explanation:

a) In this system the initial energy is elastic potential energy that when the toy is released is transformed into gravitational potential energy and kinetic energy,

Starting point. Maximum compression

       Em₀ = = ½ k x²

where x is the compression of the spring

Final point

        = K + U = ½ m v² + m g h

as there is no friction the (total) mechanical energy is conserved

         Em₀ = Em_{f}

         ½ k x² = ½ m v² + m g h

When analyzing this expression, the total energy is always the same, what it has is a conservation

* at the half height point it is part kinetic and part gravitational potential

* at the point of maximum height it is totally gravitational potential.

b) Let's use conservation of energy

       Em₀ = ½ k x2

        = U = (m₀ + m) g h

where m₀ is the mass of the spring and m the applied mass

as there is no friction, energy is conserved

       Em₀ = Em_{f}

       ½ k x² = (m₀ + m) g h

         m₀ + m =

as the student took several values we can perform the calculations a linear graph of m Vs inverse of h  ( 1/h)

m(kg)         h(m)           1 / h (m⁻¹)

0.03          0.333         3.0

0.045        0.222        4.5

0.06          0.167         6.0

0.075        0.1333       7.5

0.093        0.111          9.0

See the attachment for the graphic,

           m = A h' - mo

where A = is the slope of the graph, h ’= 1/h and m₀ is the mass of the spring

the linear regression values for this equation are

          m = 0.0104 h' - 0.0018

we substitute

          m₀ = 0.0018 kg

          \frac{k x^{2} }{2g} = 0.0104

          k = 0.0104

let's calculate

          k = 0.0104 2 9.8 / 0.02²

          k = 509.6 N / m

i think the answer is b

i think the answer is a, hope this is right, that would be my best guess.