Answer:
a) [tex]W=69.44 J[/tex]
b) [tex]\Delta U=-69.44 J[/tex]
c) [tex]K_{f}=91.94 J[/tex]
Explanation:
a) The work is defined as the integral of the force from x₁ to x₂:
[tex]W=\int^{x_{2}}_{x_{1}}Fdx[/tex]
Here:
Now. we need to take the integral to get the work done:
[tex]W=\int^{x_{2}}_{x_{1}}(2x+4)dx[/tex]
[tex]W=x^{2}|^{x_{2}}_{x_{1}}+4x|^{x_{2}}_{x_{1}}[/tex]
[tex]W=(7^{2}-1.4^{2})+4(7-1.4)=69.44 J[/tex]
b) The change of the potential energy is minus de work, so:
[tex]\Delta U=-\Delta W[/tex]
[tex]\Delta U=-69.44 J[/tex]
c) We know that the change in kinetic energy is equal to the work done, in a conservative system, so:
[tex]W=\Delta K= 69.44 J[/tex]
And know that:
[tex]\Delta K= K_{f}-K_{i}[/tex]
We need to find the final kinetic energy, so let's solve it for Kf
[tex]K_{f}=\Delta K+K_{i}[/tex]
[tex]K_{f}=\Delta K+(1/2)mv^{2}[/tex]
[tex]K_{f}=69.44+(1/2)*5*3^{2}[/tex]
[tex]K_{f}=91.94 J[/tex]
I hope it helps you!
