the work ... 7.6N * 8.6m
... is equal to the block's KE
1/2 * 0.50 * v^2 = 7.6 * 8.6
If a 0.50-kg block initially at rest on a frictionless, horizontal surface is acted upon by a force of 7.6 N for a distance of 8.6 m, then what would be the block's velocity
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