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nataly862011 [7]
2 years ago
13

A 77.0−kg short-track ice skater is racing at a speed of 12.6 m/s when he falls down and slides across the ice into a padded wal

l that brings him to rest. Assuming that he doesn't lose any speed during the fall or while sliding across the ice, how much work is done by the wall while stopping the ice skater?
Physics
1 answer:
sukhopar [10]2 years ago
3 0

Answer:

-6112.26  J

Explanation:

The initial kinetic energy, KE_i is given by

KE_i=0.5mv_1^{2} where m is the mass of a body and v_i is the initial velocity

The final kinetic energy, KE_f is given by

KE_f=0.5mv_f^{2} where v_f is the final velocity

Change in kinetic energy, \triangle KE is given by

\triangle KE=KE_f-KE_i=0.5mv_f^{2}-0.5mv_1^{2}=0.5m(v_f^{2}-v_i^{2})

Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for v_f and 12.6 m/s for v_i and 77 Kg for m we obtain

\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26  J</u>

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