Answer:
0.11m/s
Explanation:
To solve the exercise it is necessary to apply the concepts related to the conservation of both: kinetic and spring energy(Elastic potential energy), in this way
Kinetic Energy = Elastic potential energy
[tex]KE = SE[/tex]
[tex]\frac{1}{2}mv^2 = \frac{1}{2}kX^2[/tex]
Where,
m=mass
v=velocity
k=spring constant
x=amount of compression
Re-arrange the equation to find the velocity we have,
[tex]v^2 = \frac{kX^2}{m}[/tex]
[tex]v^2 = \frac{10(0.25)^2}{50}[/tex]
[tex]v = \sqrt{0.0125}[/tex]
[tex]v = 0.11m/s[/tex]
Therefore the maximum speed of the bananas is 0.11m/s
