Answer:
a) U(x)=30x^2 +6x^3
b) v1=7.9 m/s
Explanation:
a) the potential energy function is equal to:
[tex]U(x)=-\int\limits^ {}F(x) \, dx =\alpha (\frac{x^{2} }{2})+\beta (\frac{x^{3} }{3})+c[/tex]
when U(x)=0 and x=0
[tex]U(x)=\alpha (0)+\beta (0)+c[/tex]
c=0
substituting the values we have to:
[tex]U(x)=60*(\frac{x^{2} }{2})+18*(\frac{x^{3} }{3})+0=30x^{2} +6x^{3}[/tex]
b) the speed of the object is equal to:
[tex]Ek_{1}+U_{1}=Ek_{2}+U_{2}[/tex]
[tex]\frac{1}{2}mv1^{2}+30x^{2} _{1}+6x^{3} _{1}=0+30x^{2} _{2}+6x^{3} _{2}[/tex]
if x1=1 m, x2=0.5 m and m=0.9 kg
[tex]\frac{1}{2}+30*0.5^{2}+6*0.5^{2}=30*1^{2}+6*1^{2}[/tex]
[tex]0.45v1^{2}+8.25=3.6[/tex]
Clearing v1:
v1=7.9 m/s
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
