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The only force acting on an object moving along the x-axis is the conservative force given by F(x) = (2.00 N/m)x + (1.00 N/m3)x3. (a) What is the change in potential energy when the object moves from x = 1.00 m to x = 2.00 m? (b) What is the change in kinetic energy when the object moves from x = 1.00 m to x = 2.00 m?

Respuesta :

Answer:

Explanation:

Given

Force [tex]F(x)=2x+x^3[/tex]

Change in Potential Energy from x=1 to x=2 m is given by

[tex]\Delta U=\int F(x)dx[/tex]

[tex]\Delta U=\int_{1}^{2}F(x)dx[/tex]

[tex]\Delta U=\int_{1}^{2}(2x+x^3)dx[/tex]

[tex]\Delta U=|x^2+\frac{x^4}{4}|_1^{2}[/tex]

[tex]\Delta U=6.75 J[/tex]

(b)Change in Kinetic Energy

If the Force is conservative in nature then the total change in mechanical energy is zero thus

Thus [tex]\Delta K.E.=U_1-U_2=-\Delta U=-6.75 J[/tex]

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