A 0.324kg air track glider moves linearly with an initial speed of 1.37m/s. In o

Question

A 0.324kg air track glider moves linearly with an initial speed of 1.37m/s. In order to increase the speed of the mass, 7.31J of work are done. What is the final speed of the glider?

answer:
Problem 2:
1J = 1 Kg*m^2/s^2
(1/2)m v(initial)^2 + work done to system = (1/2) m v(final)^2
(1/2)(0.324kg)(1.37m/s)^2 + 7.31 kg*m^2/s^2 = (1/2)(0.324 lg) v(final)^2
V(final) = 6.86 m/s

why is that he equates 1/2m vfinal^2 to 1/2m vinitial^2 + work done and why is that he adds the work done? I dont understand. please help?

Explanation / Answer

the mass of air track glider is m = 0.324 kg

the initial speed is u = 1.37 m/s

the work done is W = 7.31 J

let v be the final speed of glider

according to work energy theorem we have

W = (1/2)m x (v^2 - u^2)

or v^2 - u^2 = (2W/m)

or v^2 = u^2 + (2W/m)

or v = (u^2 + (2W/m))^1/2

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