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

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

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

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

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

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

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

$$v_1=7.85 m/s$$