2. A2.0 kg wood box is sitting on a level steel surface. A student pushes the box and it accelerates from rest to 12.4m/s in 3.60 s. What is the net force that acting on the box? (4)

Here's how to calculate the net force acting on the wooden box:<br /><br />1. **Find the acceleration:**<br /><br />Acceleration (a) is the change in velocity (Δv) divided by the change in time (Δt).<br /><br />* Δv = Final velocity - Initial velocity = 12.4 m/s - 0 m/s = 12.4 m/s<br />* Δt = 3.60 s<br /><br />a = Δv / Δt = 12.4 m/s / 3.60 s = 3.44 m/s²<br /><br />2. **Calculate the net force:**<br /><br />Newton's second law of motion states that force (F) equals mass (m) times acceleration (a).<br /><br />* m = 2.0 kg<br />* a = 3.44 m/s²<br /><br />F = m * a = 2.0 kg * 3.44 m/s² = 6.88 N<br /><br />Therefore, the net force acting on the box is approximately $\boxed{6.9 N}$<br />