A force of 123 N acts on a 10.0kg object for 45.6 meters. If the objects starts at rest, and the surface is flat and frictionless, how fast is the object moving after the force is removed.

I got acceleration to be 12.3, but I am not sure how to move on with the problem.

d = (1/2) a t^2 = 45.6

You know a, solve for t

v = a t

the work (energy) is equal to the KE of the object

f d = 1/2 m v^2

123 * 45.6 = 1/2 * 10.0 * v^2

To solve this problem, you need to use the equation of motion relating force, mass, acceleration, and displacement. The equation is:

F = m * a

Where:
F is the force applied (123 N),
m is the mass of the object (10.0 kg),
a is the acceleration of the object (unknown).

From your calculations, you found that the acceleration of the object is 12.3 m/s^2. Now, you can use this value to calculate the final velocity (Vf) of the object after the force is removed.

To find the final velocity, you can use the equation of motion relating initial velocity (Vi), final velocity (Vf), acceleration (a), and displacement (s):

Vf^2 = Vi^2 + 2 * a * s

Since the object starts at rest (Vi = 0 m/s), the equation simplifies to:

Vf^2 = 2 * a * s

Plug in the values:

Vf^2 = 2 * 12.3 m/s^2 * 45.6 m

Now, solve for Vf by taking the square root of both sides of the equation:

Vf = √(2 * 12.3 m/s^2 * 45.6 m)

Vf ≈ 17.055 m/s

Therefore, the object is moving at approximately 17.055 m/s after the force is removed.