Question

At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 32.8105 cm on a spring with a spring constant of 13.9602 N/m. The mass of the bananas is 49.0943 kg. What is the maximum speed of the bananas?

Answer 1

Answer:

0.175 m/s

Explanation:

32.8105 cm = 0.328105 m

According to the law of conservation in energy, as the spring oscillates, its kinetic energy is converted to spring elastic energy and vice versa. So the maximum speed of the bananas occur when spring elastic energy is 0 and vice versa, spring elastic energy is maximum (at amplitude) with kinetic energy, and speed, is 0

$E_e = E_k$

$kx^2/2 = mv^2/2$

$13.9602*(0.328105)^2/2 = 49.0943v^2/2$

$v^2 = \frac{13.9602*(0.328105)^2}{49.0943} = 0.0306$

$v = \sqrt{0.0306} = 0.175 m/s$