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 44.0 cm/s. What is the weight of the bananas in newtons?

Answer 1

Answer:

32.4289 N

Explanation:

A = Amplitude = 20 cm

$v_m$ = Maximum velocity = 44 cm/s

k = Spring constant = 16 N/m

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of object

Maximum velocity is given by

$v_m=A\omega$

Angular velocity is given by

$\omega=\sqrt{\dfrac{k}{m}}$

$v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow m=\dfrac{A^2k}{v_m^2}\\\Rightarrow m=\dfrac{0.2^2\times 16}{0.44^2}\\\Rightarrow m=3.3057\ kg$

Weight is given by

$W=mg\\\Rightarrow W=3.3057\times 9.81\\\Rightarrow W=32.4289\ N$

The weight of the bananas is 32.4289 N