Asked by Natasha

A net force of 25N is applied for 5.7s to a 12-kg box initially at rest. Waht is the speed of the box at the end of the 5.7s interval?
Answered by Henry
a = Fn/m = 25 / 12 = 2.08 m/s^2.

Va*t = 2.08m/s^2 * 5.7s. = 11.88 m/s.
Answered by Anonymous
12 m/s
Answered by donald trump
that is correct
Answered by carl
A 10-N net force is applied for 5.0 s to a 12-kg box initially at rest. What is the speed of the box at the end of the 5.0-s interval?
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