Question

a particle is subject to a force ( f(x)=2x^{2}-6x ) newtons. how much work is done on the particle if it moves from ( x = 0.6 m ) to ( x = 5.8 m )?

Explanation:

Step1: Recall the work - integral formula

The work \(W\) done by a force \(F(x)\) as a particle moves from \(x = a\) to \(x = b\) is given by \(W=\int_{a}^{b}F(x)dx\). Here, \(F(x)=2x^{2}-6x\), \(a = 0.6\), and \(b = 5.8\). So, \(W=\int_{0.6}^{5.8}(2x^{2}-6x)dx\).

Step2: Integrate term - by - term

Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have \(\int(2x^{2}-6x)dx=2\int x^{2}dx-6\int xdx\).
\(2\int x^{2}dx-6\int xdx=2\times\frac{x^{3}}{3}-6\times\frac{x^{2}}{2}=\frac{2}{3}x^{3}-3x^{2}+C\).

Step3: Evaluate the definite integral

By the fundamental theorem of calculus \(\int_{a}^{b}f(x)dx=F(b)-F(a)\), where \(F(x)=\frac{2}{3}x^{3}-3x^{2}\).
\(F(5.8)=\frac{2}{3}(5.8)^{3}-3(5.8)^{2}=\frac{2}{3}\times195.112-3\times33.64=\frac{390.224}{3}-100.92\approx130.075 - 100.92=29.155\).
\(F(0.6)=\frac{2}{3}(0.6)^{3}-3(0.6)^{2}=\frac{2}{3}\times0.216-3\times0.36 = 0.144 - 1.08=-0.936\).
\(W=F(5.8)-F(0.6)=29.155-(-0.936)=30.091\).

Answer:

\(30.09\) joules