A certain spring is found not to obey Hooke’s law; it exerts a restoring force Fx(x)=−αx−βx2 if it is stretched or compressed, where α=60.0N/m and β=18.0N/m2. The mass of the spring is negligible. (a) Calculate the potential-energy function U(x) for this spring. Let U = 0 when x = 0. (b) An object with mass 0.900 kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 m to the right (the +x-direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 m to the right of the x = 0 equilibrium position?

Respuesta :

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

V^2=8.28m/s

Explanation:The law of conservation of energy is given by K1+U1= K2+U2 ...eq 1

Kinetic energy K.E= 1/2 mv^2

Restoring force function F(x)= -60x - 18x^2

But F(x)= -dU/dx

dU(x)=-F(x)dx

Integrating U(x)= -integral F(x)dx + U(0)

Substituting, we get

U(x) = - integral(-60x-18x^2)dx+U(0)

U(x)= 30x^2+6x^3+U(0)

U=0 at x=0

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

b) Given : x1=1.00m,x2= 0.50m ,V1=0, V2=?

Substituting into eq (a)

U1= 30(1.00)^2+6(1.00)^3=36J

Using x2=0.5 into eq(a)

U2=30(0.50)^2+6(0.50)^3=8.25J

Object at rest K1=0

0+36=K2+8.25

K2=27.75J

Given; m =0.900kg, V2=?

27.75=1/2×0.900×V2^2

V2= SQRT(2×27.75)/0.81

V2= 8.28m/s

The relationship between force, potential energy and energy conservation allows to find the results for the questions about the spring are:

   a) The potential energy is: U = 30 x² + 6 x³

   b) The velocity is: v = 7.85 m / s

Given parameters.

  • Restorative force f = - α x - β x²
  • Constants values  α = 60.0 N / m and β = 18.0 N / m²
  • Body mass m = 0.900 kg
  • Displacement initial x₁ = 1,0 m and final x₂ = 0,5 m

To find

    a) Potential energy.

    b) Speed.

a) Force and potential energy are related by the expression.

         [tex]F = - \frac{dU}{dx}[/tex]  

Where F is the force and U the potential energy.

       

         dU = -  F dx

         ∫ dU = - ∫∫ (-α x - β x²) dx

        U- U₀ = [tex]\alpha \frac{x^2}{2} + \beta \frac{x^3}{3}[/tex]alpha / 2 x² + beta / 3 x³

Let's substitute the constants values and indicate that U₀=0  when x=0.

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

b) They ask the speed of the block between two points, as they indicate that there is no friction we can use the theorem of conservation of mechanical energy, which states that energy is conserved at all points.

Starting point.

     Em₀ = U (1)

Final point.

     [tex]Em_f[/tex] = K + U (0.5)

Energy is conserved.

     Em₀ = Em_f

     U (1) = K + U (0.5)

Where the kinetic energy is

    K = ½ m v²

Let's substitute.

      v² = [tex]\frac{2}{m} [ U(1) - U(0.5)][/tex]  

Let's calculate.

      v² = [tex]\frac{2}{0.900}[/tex] [ (30 1² + 6 1³) - (30 0.5² + 6 0.5³) ]  

      v = [tex]\sqrt{\frac{2 \ 27.75 }{ 0.900} }[/tex]

      v = 7.85 m / s

In conclusion using the relationship between force, energy potential and the conservation of energy we can find the results for the questions about the spring are:

   a) The potential energy is: U = 30 x² + 6 x³

   b) The velocity is: v = 7.85 m / s

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