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3 years ago

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At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 34.7998 cm on a spring with a sprin

g constant of 13.5666 N/m. It is observed that the maximum speed of the bunch of bananas is 70.8966 cm/s. The acceleration of gravity is 9.8 m/s^2 .
1. What is the weight the bananas? Answer in units of N

1 answer:

Otrada [13]3 years ago

7 0

Answer:

W = 32.03316 N

Explanation:

given,

Amplitude, A = 34.7998 cm = 0.347998 m

spring constant, k = 13.5666 N/m

maximum speed, v = 70.8966 cm/s = 0.708966 m/s

acceleration due to gravity, g = 9.8 m/s²

weight of the banana = ?

using maximum speed formula

$v_{max} = A \sqrt{\dfrac{k}{m}}$

squaring both side

$v_{max}^2 = A^2 \dfrac{k}{m}$

$m = \dfrac{A^2k}{v_{max}^2}$

$m = \dfrac{0.347998^2\times 13.5666}{0.708966^2}$

m = 3.26868 Kg

weight

W = mg

W= 3.26868 x 9.8

W = 32.03316 N

hence, weight of Banana is equal to W = 32.03316 N

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