Question

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?

Answer 1

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