Respuesta :
Answer:
ks= 133.2 N/m
Explanation:
- Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.
- This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.
- When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.
- Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.
- If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.
- So, we can write the following equation:
[tex]\Delta U + \Delta K = 0 (1)[/tex]
- where ΔU = -1/2*k*(Δx)² (2)
- and ΔK = 1/2*m*v² (3)
- Replacing in (2) and (3) by the givens, and simplifying, we can find the stiffness ks as follows:
[tex]k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)[/tex]