Question

A constant force of magnitude 10 N is applied to a body of mass 2 kg at a displacement of 4 m on a horizontal floor flat. Knowing that the body was at rest,
calculate:

a) The kinetic energy of the body

b) the final velocity of the body

Answer 1

Force(F) = 10 N , Mass(m) = 2 kg , Distance(d or s) = 4 m

a) Kinetic Energy = Work done = Force × Displacement
Kinetic Energy = F×d
Kinetic Energy (K.E.) = (10 × 4) = 40 Joules 

b) Now we have Kinetic Energy = 40 Joules
Formula of K.E. =$\frac{1}{2}$ ×mv²
Now put the values,
We get,
40=$\frac{1}{2} ×2v²$
v²=40
v=√40
v=6.32m/s
The final velocity is 6.32m/s

Answer 2

F = 10 N    => acceleration a = 10 N / 2 kg = 5 m/s^2
         u = 0 initial. 
         v^2 - u^2 = 2 a s      => v^2 = 2 * 5 * 4 = 40
         so,    v = sq root of 40.

KE = 1/2 * 2 kg * 40 = 40 Joules