An object with mass 0.900kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00m to the right (the x

Question

An object with mass 0.900kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00m to the right (the x – direction) to stretch the spring, and released. What is the speed of the object when it is 0.50m to the right of the x

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Thiên Di 1 year 2021-09-03T02:24:56+00:00 1 Answers 13 views 0

Answers ( )

    0
    2021-09-03T02:26:51+00:00

    Answer:

    7.85 m/s

    Explanation:

    We are given that

    Mass of object=m=0.900 kg

    [tex]F(x)=\alpha x-\beta x^2[/tex]

    [tex]\alpha=60 N/m[/tex]

    [tex]\beta=18N/m^2[/tex]

    [tex]F(x)=-60x-18x^2[/tex]

    U=0 when x=0

    Potential energy=[tex]-\int F(x)dx[/tex]

    Substitute the values

    [tex]U(x)=-\int (-60x-18x^2)dx[/tex]

    [tex]U(x)=60(\frac{x^2}{2})+18(\frac{x^3}{3})+C[/tex]

    Using the formula

    [tex]\int x^n dx=\frac{x^{n+1}}{n+1}+C[/tex]

    Substitute x=0

    [tex]U(0)=C\implies C=0[/tex]

    [tex]U(x)=30x^2+6x^3[/tex]

    [tex]x_1=0.5,x_2=1[/tex]

    [tex]v_2=0[/tex]

    Using law of conservation energy

    [tex]\frac{1}{2}mv^2_1+U(x_1)=\frac{1}{2}mv^2_2+U(x_2)[/tex]

    Substitute the values

    [tex]\frac{1}{2}(0.9)v^2_1+30(0.5)^2+6(0.5)^3=0+30(1)^2+6(1)^3[/tex]

    [tex]\frac{1}{2}(0.9)v^2_1+8.25=36[/tex]

    [tex]\frac{1}{2}(0.9)v^2_1=36-8.25=27.75[/tex]

    [tex]v^2_1=\frac{27.75\times 2}{0.9}[/tex]

    [tex]v_1=\sqrt{\frac{27.75\times 2}{0.9}}[/tex]

    [tex]v_1=7.85 m/s[/tex]

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