6.6 The work-energy theorem for a variable force

We are now familiar with the concepts of work and kinetic energy to prove the work-energy theorem for a variable force. We confine ourselves to one dimension. The time rate of change of kinetic energy is
\(
\begin{aligned}
\frac{\mathrm{d} K}{\mathrm{~d} t} & =\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{1}{2} m v^2\right) \\
& =m \frac{\mathrm{d} v}{\mathrm{~d} t} v \\
& =F v \text { (from Newton’s Second Law) } \\
& =F \frac{\mathrm{d} x}{\mathrm{~d} t}
\end{aligned}
\)
Thus
\(
\mathrm{d} K=F \mathrm{~d} x
\)
Integrating from the initial position \(\left(x_i\right)\) to final position \(\left(x_f\right)\), we have
\(
\int_{K_i}^{K_f} \mathrm{~d} K=\int_{x_i}^{x_f} F \mathrm{~d} x
\)
where \(K_i\) and \(K_f\) are the initial and final kinetic energies corresponding to \(x_i\) and \(x_{\mathrm{f}}\)
or
\(
K_f-K_i=\int_{x_i}^{x_f} F \mathrm{~d} x \dots(6.8a)
\)
From Eq. (6.7), it follows that
\(
K_f-K_i=W \dots(6.8b)
\)
Thus, the WE theorem is proved for a variable force.

While the WE theorem is an integral form of Newton’s second law. Newton’s second law is a relation between acceleration and force at any instant of time. Work-energy theorem involves an integral over an interval of time. In this sense, the temporal (time) information contained in the statement of Newton’s second law is ‘integrated over’ and is not available explicitly. Another observation is that Newton’s second law for two or three dimensions is in vector form whereas the work-energy theorem is in scalar form. In the scalar form, information with respect to directions contained in Newton’s second law is not present.

Example 6.9: A block of mass \(m=1 \mathrm{~kg}\), moving on a horizontal surface with speed \(v_1=2 \mathrm{~m} \mathrm{~s}^{-1}\) enters a rough patch ranging from \(x=0.10 \mathrm{~m}\) to \(x=2.01 \mathrm{~m}\). The retarding force \(F_r\) on the block in this range is inversely proportional to \(x\) over this range, \(F_r=\frac{-k}{x}\) for \(0.1<x<2.01 \mathrm{~m}\) \(=0\) for \(x<0.1 \mathrm{~m}\) and \(x>2.01 \mathrm{~m}\) where \(k=0.5 \mathrm{~J}\). What is the final kinetic energy and speed \(v_f\) of the block as it crosses this patch?

Answer: From Eq. (6.8a)
\(
\begin{aligned}
& K_f=K_t+\int_{0.1}^{2.01} \frac{(-k)}{x} \mathrm{~d} x \\
& =\frac{1}{2} m v_t^2-\left.k \ln (x)\right|_{0.1} ^{2.01} \\
& =\frac{1}{2} m v_t^2-k \ln (2.01 / 0.1) \\
& =2-0.5 \ln (20.1) \\
& =2-1.5=0.5 \mathrm{~J} \\
& v_f=\sqrt{2 K_f / m}=1 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)
Here, note that \(\ln\) is a symbol for the natural logarithm to the base \(e\) and not the logarithm to the base \(10\left[\ln X=\log _e X=2.303 \log _{10} X\right]\).