   Chapter 13, Problem 19P

Chapter
Section
Textbook Problem

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

To determine
The weight of the bananas in Newton.

Explanation

Given info: The spring has a spring constant 16.0Nm-1 . The amplitude is 20 cm. The maximum speed of the bananas is 40cms-1 .

Explanation:

The total energy of the object-spring system at any instance is given by:

E=KE+PEs

Here,

E is the total energy of the system

KE is the kinetic energy of the system

PEs is the elastic potential energy of the system

The kinetic energy of the system is given by:

KE=12mv2

Here,

m is the mass of the system

v is the velocity of the system

The elastic potential of the system is given by:

PEs=12kx2

Here,

k is the force constant of the spring

x is the displacement of the spring from its equilibrium position

At the equilibrium position the total energy is stores as kinetic energy and at the turning points the total energy is stored as elastic potential energy.

Hence using conservation of energy

12mvmax2=12kA2

• vmax is the maximum velocity of the bananas
• A is the amplitude of the oscillation

On re-arranging

m=kA2vmax2

Substituting 16.0Nm-1 for k , 20 cm for A , 40cms-1 for vmax to find mass m

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