Answer with Explanation:
Mass of block=1.1 kg
Th force applied on block is given by
F(x)=[tex](2.4-x^2)\hat{i}N[/tex]
Initial position of the block=x=0
Initial velocity of block=[tex]v_i=0[/tex]
a.We have to find the kinetic energy of the block when it passes through x=2.0 m.
Initial kinetic energy=[tex]K_i=\frac{1}{2}mv^2_i=\frac{1}{2}(1.1)(0)=0[/tex]
Work energy theorem:
[tex]K_f-K_i=W[/tex]
Where [tex]K_f=[/tex]Final kinetic energy
[tex]K_i[/tex]=Initial kinetic energy
[tex]W=Total work done[/tex]
Substitute the values then we get
[tex]K_f-0=\int_{0}^{2}F(x)dx[/tex]
Because work done=[tex]Force\times displacement[/tex]
[tex]K_f=\int_{0}^{2}(2.4-x^2)dx[/tex]
[tex]K_f=[2.4x-\frac{x^3}{3}]^{2}_{0}[/tex]
[tex]K_f=2.4(2)-\frac{8}{3}=2.13 J[/tex]
Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J
b.Kinetic energy =[tex]K=2.4x-\frac{x^3}{3}[/tex]
When the kinetic energy is maximum then [tex]\frac{dK}{dx}=0[/tex]
[tex]\frac{d(2.4x-\frac{x^3}{3})}{dx}=0[/tex]
[tex]2.4-x^2=0[/tex]
[tex]x^2=2.4[/tex]
[tex]x=\pm\sqrt{2.4}[/tex]
[tex]\frac{d^2K}{dx^2}=-2x[/tex]
Substitute x=[tex]\sqrt{2.4}[/tex]
[tex]\frac{d^2K}{dx^2}=-2\sqrt{2.4}<0[/tex]
Substitute x=[tex]-\sqrt{2.4}[/tex]
[tex]\frac{d^2K}{dx^2}=2\sqrt{2.4}>0[/tex]
Hence, the kinetic energy is maximum at x=[tex]\sqrt{2.4}[/tex]
Again by work energy theorem , the maximum kinetic energy of the block between x=0 and x=2.0 m is given by
[tex]K_f-0=\int_{0}^{\sqrt{2.4}}(2.4-x^2)dx[/tex]
[tex]k_f=[2.4x-\frac{x^3}{3}]^{\sqrt{2.4}}_{0}[/tex]
[tex]K_f=2.4(\sqrt{2.4})-\frac{(\sqrt{2.4})^3}{3}=2.48 J[/tex]
Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J
