Answer to Question #197890 in Mechanics | Relativity for DANI

Question #197890

1)A single conservative force acting on a particle within a system varies as ๐นโƒ— = (โˆ’๐ด๐‘ฅ + ๐ต๐‘ฅ

2

)๐‘–, ฬ‚

where A and B are constants, ๐นโƒ— is in newtons, and x is in meters. (a) Calculate the potential 

energy function U(x) associated with this force for the system, taking U=0 at x=0. Find (b)

the change in potential energy and (c) the change in kinetic energy of the system as the 

particle moves from x=2m to x-3m.


1
Expert's answer
2021-05-24T15:51:15-0400

Expression of conservative force, Fโƒ—=(โˆ’Ax+Bx2)i^ N\vec F=(-Ax+Bx^2)\hat i\space N

(a) Potential energy,

dU=Fโƒ—.dxd U=\vec F.dx

dU=(โˆ’Ax+Bx2)dxd U=(-Ax+Bx^2)dx


At x = 0, U = 0 (given)

โˆซ0UdU=โˆซ0x(โˆ’Ax+Bx2)dx\int_0^U d U=\int_0^x(-Ax+Bx^2)dx

U(x)=โˆ’Ax22+Bx33U(x)=\dfrac{-Ax^2}{2}+\dfrac{Bx^3}{3}


(b) Change in potential energy as x changes from 2 to 3

U(3)โˆ’U(2)=โˆฃโˆ’Ax22+Bx33โˆฃ23U(3)-U(2)=\bigg|\dfrac{-Ax^2}{2}+\dfrac{Bx^3}{3}\bigg|_2^3

ฮ”U=โˆ’9A2+27B3+4A2โˆ’8B3\Delta U=\dfrac{-9A}{2}+\dfrac{27B}{3}+\dfrac{4A}{2}-\dfrac{8B}{3}

ฮ”U=โˆ’5A2+19B3\Delta U=\dfrac{-5A}{2}+\dfrac{19B}{3}


(c)  If we consider the particle alone as a system, the change in its kinetic energy is the work done by the force on the particle:

W=ฮ”KW=ฮ”K

For the entire system of which this particle is a member, this work is internal work and equal to the negative of the change in potential energy of the system:

ฮ”K=โˆ’ฮ”U=2.5Aโˆ’6.33Bฮ”K=โˆ’ฮ”U=2.5A-6.33B


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment