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

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

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

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

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

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

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

$v_1=7.85 m/s$