Answer:
U = 0.5 * k *(x + d - h_max)^2 + m*g*h_max
Explanation:
Given:
- The extension in spring @ equilibrium = x m
- The spring constant = k
- The amount of distance pulled down = d
- mass of the toy = m
Find:
- The total mechanical energy E_top at the top position h_max in terms of the available variables.
Solution:
- First we need to determine the types of Energy that are in play:
- The Elastic potential Energy E_p in a spring is given:
E_p: 0.5 * k * (ext)
- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:
ext = Equilibrium + distance pulled - h_max = (x + d - h_max)
Hence, the elastic potential energy will be:
E_p = 0.5 * k *(x + d - h_max)^2
- The gravitational potential energy E_g is given by:
E_g = m*g*h_max
Where, bottom most position is taken as reference (datum).
- The kinetic Energy E_k is given by:
E_k = 0.5*m*v_top^2
- Since we know that the maximum height is reached when velocity is zero
Hence, E_k = 0.5*m*0^2 = 0.
The total Energy of the system U is sum of all energies and play:
U = E_p + E_k + E_g
U = 0.5 * k *(x + d - h_max)^2 + m*g*h_max
or (-0.6, 0.8)
