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

0.11m/s

Step-by-step 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`

`(1)/(2)mv^2 = (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 = (kX^2)/(m)`

`v^2 = (10(0.25)^2)/(50)`

`v = √(0.0125)`

`v = 0.11m/s`

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