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What average force is needed to accelerate a 7.00-gram pellet from rest to 125 m/s over a distance of 0.800m along the barrel of a rifle. I tried this and I got 754.88 Answer: 76.84 N
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The change in kinetic energy must equal the work done. It starts at rest, hence the change in KE is just the ending KE, and if that is equal to work then \[ \frac{1}{2}mv^2 = Fd \] hence \[ d = \frac{mv^2}{2d} \] where m = 0.007 kg v = 125 m/s d = 0.8 m
**correction F = ...
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