A single conservative force acting on a particle varies as = (-Ax + Bx2) N, wher

Question

A single conservative force acting on a particle varies as = (-Ax + Bx2) N, where A and B are constants and x is in meters. Accurately round coefficients to three significant figures.

(a) Calculate the potential energy function U(x) associated with this force, taking U = 0 at x = 0. (Use any variable or symbol stated above as necessary.)

U= A(X^2/2) - B(X^3/3)

(b) Find the change in potential energy and change in kinetic energy as the particle moves from x = 1.80 m to x = 3.40 m. (Use any variable or symbol stated above as necessary.)

?U =

?K =

Explanation / Answer

Force F= (-Ax + Bx^2) N a) Work Done (stored as PE) in elemental movement dx =F*dx J U(x) In Lim dx ? 0 ? F*dx = ? (-Ax + Bx^2)*dx for range 0 to x. U(x) = [ -Ax^2/2 + Bx^3/3 + C ] for range 0 to x. U(0) = 0 = C U(x) = [ -Ax^2/2 + Bx^3/3 ] J b) ?U= [ -A*3.4^2/2 + B*3.4^3/3 ] - [ -A*1.8^2/2 + B*1.8^3/3 ] = -A/2*[3.4^2 - 1.8^2] + B/3*[3.4^3 - 1.8^3] J

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