This problem can be solved via the work-energy theorem. First, we'll break down the applied force into its components because the force is not being applied horizontally, but at an angle of 30 degrees below the horizontal. We can use the cosine of this angle to find the horizontal component of the force, since cos(θ) = Adjacent/Hypotenuse.
Hence, the horizontal component of the force = 40.0 N * cos(30.0 degrees) = 34.64101615137755 N.
The work done by this force component is equal to the force times the distance over which the force is applied. So we calculate the work as:
Work done = force_horizontal * distance = 34.64101615137755 N * 16.0 m = 554.2562584220408 J.
According to the work-energy theorem, the work done on the object is equal to the change in its kinetic energy. Since the object starts from rest, all of this work goes into changes in kinetic energy. Hence, the final kinetic energy of the object = work done = 554.2562584220408 J.
Finally, we calculate the final speed of the object using the formula for kinetic energy 1/2 * mass * final speed^2:
Rearranging this for speed, we get final speed = sqrt(2 * final kinetic energy / mass). Substituting the known values we have, the final speed = sqrt(2 * 554.2562584220408 J / 2.0 kg) = 23.542647651061696 m/s.
So, the speed of the 2.0-kg object, which is pushed for 16.0 m by a 40.0 N force directed 30 ⋅ below the horizontal on a level, frictionless floor, will be 23.54 m/s.