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A 75 gFrisbee is thrown from a point 1.1 mabove the ground with a speed of 12 m/s.When it has reached a height of 2.1 m, its speed is 10.5 m/s. What was the reduction in Emec of the Frisbee-Earth system because of air drag?

Short Answer

Expert verified

The amount of decrease in mechanical energy of the Frisbee-Earth system is Emec=0.53J.

Step by step solution

01

Step 1: Given Data

The mass of a Frisbee ism=0.075kg

The initial speed of Frisbee,v1=12m/s

The final speed of Frisbee,v2=10.5m/s

The height of Frisbee,h1=1.1m

The final height of Frisbee,role="math" localid="1661232990159" h2=2.1m

02

Determining the concept

Use the equation of change in mechanical energy to solve this problem. Mechanical energy is the sum of kinetic and potential energy

Formulae are as follow:

Emec=K1+U1-K2+U2K=12mv2U=mgh

where, K is kinetic energy, Uis potential energy, m is mass, v is velocity, g is an acceleration due to gravity, h is height andEmec is mechanical energy.

03

Determining the amount of decrease in mechanical energy of the Frisbee-Earth system

The equation for change in mechanical energy is,

Emec=K1+U1-K2+U2

Using K=12mv2and U=mgh,

Emec=12mv12-v22+mgh1-h2Emec=120.075122-10.52+0.0759.811.1-2.1Emec=1.27-0.74Emec=0.53J

Therefore, the amount of decrease in mechanical energy of the Frisbee-Earth system is Emec=0.53J.

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