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Chapter 6: Q. 66 (page 157)

A 60kgskater is gliding across frictionless ice at 4.0m/s. Air resistance is not negligible. You can model the skater as a 170鈭�cm-tall,36鈭�cm-diameter cylinder. What is the skater's speed2.0s later?

Short Answer

Expert verified

Speed of the skater after2Sec is3.4m/s.

Step by step solution

01

:Given.

Mass of the skater, m=60kg.

Speed of the skater, v=4m/s.

Diameter of the skater, d=36cm.

Height of the skater,h=170cm.

02

Formula used.

The air friction is given by the formula:

F=12蚁脳A脳v2

Here,

蚁is the density of air.

Ais the surface area of the cylinder.

vis the velocity of the skater.

03

Calculation.

The surface area of the skater is calculated as:A=蟺dH

Plugging the values in the above equation,

A=蟺脳0.36脳1.7=1.92m2.

Now drag force due to air is calculated as:

F=12脳1.2脳1.92脳42=18.43N.

The deceleration on the skater due to air friction is calculated as:

F=ma18.43=60脳aa=0.30m/s2.

The speed of the skater after 2sec is calculated as:

role="math" localid="1647774441411" v=u鈭�at=4鈭�0.3脳2=3.4m/s.

04

Conclusion.

Speed of the skater after 2sec is3.4m/s.

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