Question

At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 25 cm on a spring with a spring constant of 10 N/m. The mass of the bananas is 50 kg.What is the maximum speed of the bananas?

Answer 1

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

$KE = SE$

$\frac{1}{2}mv^2 = \frac{1}{2}kX^2$

Where,

m=mass

v=velocity

k=spring constant

x=amount of compression

Re-arrange the equation to find the velocity we have,

$v^2 = \frac{kX^2}{m}$

$v^2 = \frac{10(0.25)^2}{50}$

$v = \sqrt{0.0125}$

$v = 0.11m/s$

Therefore the maximum speed of the bananas is 0.11m/s