Question

A certain spring found not to obey Hooke's law exerts a restoring force Fx(x) = —ax — /3x2 if it is stretched or compressed, where a = 60.0 Him and /3 = 18.0 HIM2. The mass of the spring is negligible. (a) Calculate the potential-energy func-tion 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 at-tached 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:

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

b) v1=7.9 m/s

Explanation:

a) the potential energy function is equal to:

$U(x)=-\int\limits^ {}F(x) \, dx =\alpha (\frac{x^{2} }{2})+\beta (\frac{x^{3} }{3})+c$

when U(x)=0 and x=0

$U(x)=\alpha (0)+\beta (0)+c$

c=0

substituting the values ​​we have to:

$U(x)=60*(\frac{x^{2} }{2})+18*(\frac{x^{3} }{3})+0=30x^{2} +6x^{3}$

b) the speed of the object is equal to:

$Ek_{1}+U_{1}=Ek_{2}+U_{2}$

$\frac{1}{2}mv1^{2}+30x^{2} _{1}+6x^{3} _{1}=0+30x^{2} _{2}+6x^{3} _{2}$

if x1=1 m, x2=0.5 m and m=0.9 kg

$\frac{1}{2}+30*0.5^{2}+6*0.5^{2}=30*1^{2}+6*1^{2}$

$0.45v1^{2}+8.25=3.6$

Clearing v1:

v1=7.9 m/s

Answer 2

Answer:

a)U( x)=30x^²+6x^³

b)V₂=7.85m/s

Explanation:

(a)CALCULATING THE POTENTIAL ENERGY

recall law of conservation of matter which states

K+U= K+U...................................................equ.(*)

where K₁ is the initial kinetic energy

U₁ is the initial potential energy

but Kinetic Energy (K=¹/₂mv²)......................equ.(**)

restoring force formular F(x)= -60x-18x²

getting the derivatives of the conservative force,

then F(x)= -du(x)/dx

du(x)= -F(x)dx

introducing the integral sign, by integrating

U(x)= -∫(-60x-18x²)dx + U(0)

integrating

∴ U(x)= 30x²+6x³ + U(0)......................eqn. (a)

when U=0 at x=0 substituting in equation ( a)

then

=30x² ⁺6x³ + U(0).................................eqn. (b)

when x₁= 1m v₁=0

x₂=0.5m v₂= unknown

substituting x₁= 1m and v₁=0 into eqn (b)

U₁= 30(1)² + 6(1)³= 36J

substituting x₁= 0.5m and v₁=0 into eqn (b)

U₂=30(0.5)²+6(0.5)³=8.25J

(b) CALCULATING THE SPEED OF THE OBJECT

from equ. (*) when the object is from rest i.e k₁=0

0+36 = k₂ +8.25

making k₂ subject of formular

k₂=27.75J

then when m= 0.90kg, k₂=27.75 substitute into equ. (**)

27.75=¹/₂(0.90)v²₂

v₂=7.85m/s