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

32.4289 N

Step-by-step 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{(k)/(m)}`

`v_m=A\sqrt{(k)/(m)}\\\Rightarrow m=(A^2k)/(v_m^2)\\\Rightarrow m=(0.2^2* 16)/(0.44^2)\\\Rightarrow m=3.3057\ kg`

Weight is given by

`W=mg\\\Rightarrow W=3.3057* 9.81\\\Rightarrow W=32.4289\ N`

The weight of the bananas is 32.4289 N