Answer:
The work done by the force on the particle is 29.85 J.
Explanation:
The work is given by:
[tex] W = ^{x_{2}}_{x_{1}}\int F_{x} dx [/tex]
Where:
x₁: is the lower limit = 0 m
x₂: is the upper limit = 1.9 m
Fₓ: is the force in the horizontal direction = (4.5 + 13.7x - 1.5x²)N
[tex]W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx[/tex]
[tex] W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0} [/tex]
[tex] W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3} [/tex]
[tex] W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3} [/tex]
[tex]W = 29.85 J[/tex]
Therefore, the work done by the force on the particle is 29.85 J.
I hope it helps you!
