Question

To understand the meaning and possible applications of the work-energy theorem. In this problem, you will use your prior knowledge to derive one of the most important relationships in mechanics: the work-energy theorem. We will start with a special case: a particle of mass m moving in the x direction at constant acceleration a. During a certain interval of time, the particle accelerates from vi to vf, undergoing displacement s given by s=xf−xi.

Answer 1

Explanation:

The particle travels between two positions $x_i$ and $x_f$, this can be considered as the result of adding infinite elementary displacements. Therefore, the total work done by force in this displacement is given by:

$W=\int\limits^{x_f}_{x_i} {F\cdot} \, dx$

According to Newton's second law:

$F=ma=m\frac{dv}{dt}$

Recall that $\frac{dx}{dt}=v$ and $K=\frac{mv^2}{2}$. So, we have:

$W=\int\limits^{x_f}_{x_i} {ma\cdot} \, dx\\W=m\int\limits^{x_f}_{x_i} {\frac{dv}{dt}\cdot} \, dx\\W=m\int\limits^{v_f}_{v_i} {v\cdot} \, dv\\\\W=m[\frac{v_f^2}{2}-\frac{v_i^2}{2}]\\\\W=K_f-K_i\\W=\Delta K$