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 4 years 2021-09-03T02:24:56+00:00 1 Answers 23 views 0

Answers ( )

  1. Neala
    0
    2021-09-03T02:26:51+00:00
    Reply

    Answer:

    7.85 m/s

    Explanation:

    We are given that

    Mass of object=m=0.900 kg

    F(x)=\alpha x-\beta x^2

    \alpha=60 N/m

    \beta=18N/m^2

    F(x)=-60x-18x^2

    U=0 when x=0

    Potential energy=-\int F(x)dx

    Substitute the values

    U(x)=-\int (-60x-18x^2)dx

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

    Using the formula

    \int x^n dx=\frac{x^{n+1}}{n+1}+C

    Substitute x=0

    U(0)=C\implies C=0

    U(x)=30x^2+6x^3

    x_1=0.5,x_2=1

    v_2=0

    Using law of conservation energy

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

    Substitute the values

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

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

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

    v^2_1=\frac{27.75\times 2}{0.9}

    v_1=\sqrt{\frac{27.75\times 2}{0.9}}

    v_1=7.85 m/s

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