a = Fn/m = 25 / 12 = 2.08 m/s^2.
Va*t = 2.08m/s^2 * 5.7s. = 11.88 m/s.
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?
4 answers
12 m/s
that is correct
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?