K.E=1 upon 2  m  v square. 
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is it ok,coz i could't type what i knew
Consider an object of mass m moving with uniform velocity u. Let a constant force F act on it so that it is displaced through a distance s. Then, a work W is done on it so that the velocity changes from u to v. From the equation of motion, v^2 - u^2 = 2aS ... S = v^2 - u^2/2a From the second law of motion, we know that F = ma. W = F X S = ma X v^2 - u^2/2a (cancel both 'a') = m(v^2 - u^2/2) ...W = 1/2m(v^2 - u^2) If the object is at rest initially, then u = 0 So, W = 1/2 mv^2 This work is stored as Kinetic Energy and therefore, K.E = 1/2 mv^2 ! (Where ^2 is coming that means square, 2 comes abover the letter AMD wherever / is coming it means division) hope it helps! ✌️
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