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

Answer 1

7.85 m/s

Step-by-step 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((x^2)/(2))+18((x^3)/(3))+C`

Using the formula

`\int x^n dx=(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

`(1)/(2)mv^2_1+U(x_1)=(1)/(2)mv^2_2+U(x_2)`

Substitute the values

`(1)/(2)(0.9)v^2_1+30(0.5)^2+6(0.5)^3=0+30(1)^2+6(1)^3`

`(1)/(2)(0.9)v^2_1+8.25=36`

`(1)/(2)(0.9)v^2_1=36-8.25=27.75`

`v^2_1=(27.75* 2)/(0.9)`

`v_1=\sqrt{(27.75* 2)/(0.9)}`

`v_1=7.85 m/s`