Asked by fatimah
If a 0.50-kg block initially at rest on a frictionless, horizontal surface is acted upon by a force of 5.7 N for a distance of 8.6 m, then what would be the block's velocity?
Answers
Answered by
Damon
(1/2) m v^2 = 5.7 * 8.6
v^2 = 2 * 5.7 * 8.6 / 0.50
v^2 = 2 * 5.7 * 8.6 / 0.50
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