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Chapter 1: Q18E (page 264)

Two figure skaters, one weighing 625 N and the other 725 N, push off against each other on frictionless ice. (a) If the heavier skater travels at 1.50 m/s, how fast will the lighter one travel? (b) How much kinetic energy is 鈥渃reated鈥� during the skaters maneuver, and where does this energy come from?

Short Answer

Expert verified

Thus, (a) the speed of a lighter skater is -1.75 m/s.

(b) the kinetic energy is 180 J.

Step by step solution

01

Given in the question.

The weight of one skater isF1=625N

The weight of other skaters isF2=725N

Speed of heavier skater is V1=150m/s.

Speed of lighter skater is V2.

02

(a) Speed of lighter skater.

Use the conservation of momentum to obtain the speed of a lighter skater:

0=m1v1+m2v2v2=-v1m1m2=-150625725=-1.75m/s

The negative means that the other one moves in the opposite direction.

Hence, the speed of a lighter skater is -1.75 m/s.

03

(b) Kinetic energy created the skaters.

The energy is calculated as follows:

E=12m1v1+m2v2=12F1gv1+F2gv2=12625101.50+725101.75=180J

This energy comes from their muscles, which receive the energy from the chemical reactions in the body.

Hence, the kinetic energy is 180 J.

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