Question

(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroid, travelling at a speed of 3 m/s. The payload has a mass of 29 kg. If the spring is to be compressed initially an amount of 1.4 m, what stiffness ks must the spring be designed to have

Answer 1

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:

       $\Delta U + \Delta K = 0 (1)$

• 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:

       $k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)$