Answer:
The answer is: 1913.5158 J
Explanation:
First we need to derivate the displacement equation:
[tex]x = 1.1+1.8t-3.1t^{2} +1.1t^{3} \\\frac{dx}{dt} = 1.8-6.2t+3.3t^{2} \\\frac{d^{2} x}{dt^{2} } = -6.2+6.6t \\[/tex]
Second, we need to know the 2nd Law of Newton:
[tex]F = m*a = m\frac{d^{2} x}{dt^{2} }[/tex]
So, we have:
[tex]F = 2.6*(-6.2+6.6t)[/tex]
Third, to calculate the work done on the particle we need the next equation:
[tex]W= \int\limits^a_b {F} \, dx[/tex]
Then, we have everything to replace in the variables(dx is replaced by the first derivative) :
[tex]W= \int\limits^{4.4}_{0} {2.6*(-6.2+6.6t)*(1.8-6.2t+3.3t^{2})} \, dt[/tex]
And finally we have as a result:
W = 1913.5158 J
