Answer:
ugmd = 1/2 kx²
d = (1/2 kx²) / (ugm)
= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)
= 7.4 m
ugmd = 1/2 mv²
v = √2ugd
= √(2(0.23)(9.81 m/s²)(7.4 m)
= 5.8 m/s
Explanation:
Work done by a given force is given by

here on sled two forces will do work
1. Applied force by Max
2. Frictional force due to ground
Now by force diagram of sled we can see the angle of force and displacement
work done by Max = 

Now similarly work done by frictional force



Now total work done on sled


Well you need to have lots of heat
At the end of the laps, the runner's displacement is zero.
Answer:
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Explanation:
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