Dynamics: Work energy theorem

1.Simple case: constant acceleration and straight line

We consider the special case of a moving object along a straight line, say x-axis, with a constant net force acting on it (then constant acceleration according to Newton's second law). As the object moves from x_i to x_f, its speed changes from v_i to v_f. From the formula in Kinematics, we have:

2a_x(x_f - x_i) = v_f² - v_i²

2a_x(x_f - x_i) = v_f² - v_i²     (1)

a_x = ΣF_x/m is the acceleration component along the x-axis.
The net work along the path from x_i to x_f is:
W_net = Σ F_x (x_f - x_i)     (1)
Since a_x is constant, we have:
W_net = m Σ a_x (x_f - x_i)
By multiplying by m both sides of the first relationship (1), it becomes:
m a_x(x_f - x_i) = m (1/2)[v_f² - v_i²]
That is:
W_net = (1/2)m[v_f² - v_i²] = KE_f - KE_i

This result is called the work energy theorem.

W_net is the work done by the net force on the moving object of mass m from the initial state with speed v_i to the final state with a speed v_f.

Notice that the net work is the work of the net force or the contribution of the work of each individual force.

2. General case: derivation of the formula

Newton's second law states that the acceleration of a moving particle is determined by the net force acting on it, that is :

ΣF_ext = F_net = m a

Let's form the dot product of each side of this equation with an infinitesimal displacement dr along the path taken by the object.
dr . ΣF_net = m dr.a
Σdr.F_net = m dr.a
Σdr.F_net = dW
dr/dt = v, so
m dr.a = m (v dt) .dv/dt = m v dv
Therefore:
dW = m v dv
∫ dW = ∫ m v dv = m [(1/2)v²]
From v_i to _f

Thus:
W = (1/2) m [v_f² - v_i²]