Question

At an outdoor market, a bunch of bananas attached to the bottom of a vertical spring of force constant 16.0 N/m is set into oscillatory motion with an amplitude of 20.0 cm. It is observed that the maximum speed of the bunch of bananas is 37.6 cm/s. What is the weight of the bananas in newtons

Answer 1

Answer:

Explanation:

Answer:

4.53 kg

Explanation:

spring constant, K = 16 N/m

Amplitude, A = 20 cm = 0.2 m

Maximum speed, v = 37.6 cm/s = 0.376 m/s

The formula for maximum speed is given by

v = ωA

where, ω is the angular frequency

0.376 = ω x 0.2

ω = 1.88 rad/s

$\omega =\sqrt{\frac{K}{m}}$

$m = \frac{K}{\omega ^{2}}$

$m = \frac{16}{1.88 ^{2}}$

m = 4.53 kg

Thus, the mass of banana bunch is 4.53 kg.