Question

9. A net force of 25 N is applied for 5.7 s to a 12-kg box initially at rest. What is the speed of the box at the end of the 5.7-s interval?A) 1.8 m/sB) 12 m/sC) 3.0 m/sD) 7.5 m/sE) 30 m/s

Answer 1

11.875m/s

Step-by-step explanation:

F=ma a= (v) /t

f=m(v) /t

25N = 12kg v /5.7s

v= (25N×5.7s) / 12kg

v= 11.875m/s