How do you work out Kinetic Energy without being given Velocity (Mechanical Principles)

I need to work out the kinetic energy of an object being pulled along a flat bench.

I know KE = 1/2 MV^2
But Im not given Velocity these are the factors I have already.
Distance Traveled = 4M
Force = 5N
Mass= 2.5 KG
Work Done = 20nm / 20J

I would really appreciate being shown the working out rather than just the answer, as I've been trying to work around this question for the past 4 hours. Thanks!

You are given work (W) and you know that work done = force times distance force acts through. You are also given force (F) and distance (d) so

W = Fd

Now work is also equal to the change in kinetic energy (DK) so

DK = W =Fd

If you assume the block started from rest then after the blcok has moved the distance d DK = kinetic energy so you can write:

KE = 1/2mv^2 = Fd

So KE = Fd plug in and you're done

are you sure it is a constant force and not a varying force? is it on a smooth surface (i.e. no friction)? and lastly, but very importantly, does it start from rest?

yes 20 J is your answer if it does start from rest.

jcherry, it is not so odd if you think about it. its just conservation of energy. you are doing work on the system by pulling it which is taking energy from your body and converting it into kinetic energy of the object.

You can get the acceleration.
F / m = a
5N / 2.5 kg = 2 m/s^2

You can get t if you are willing to accept that this started off at zero m/s.

a = 2 m/s^2
vi = 0
d = 4 m
t = ???

d = 0 + 1/2 a t^2
4 = 1/2 * 2 * t^2
t^2 = 4
t = 2 seconds.

Now almost anything will give you vf
vi = 0
t = 2 seconds
d = 4 m
vf = ???
d = (vf + vi)*t /2

4 = (vf + 0) * t/2
4 = vf*2/2
vf = 4 m/s

So the object is going from 0 to 4 m/s
The Ke = 1/2 * m * vf^2
Ke = 1/2 * 2.5 * (4)^2
Ke = 20 J which oddly is the same as the work done. So starting off at zero is the correct assumption which is why I went through all this.

a