A single conservative force acting on a particle varies as = (-Ax + Bx6) N, wher

Question

A single conservative force acting on a particle varies as = (-Ax + Bx6) 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=

(b) Find the change in potential energy and change in kinetic energy as the particle moves from x = 1.40 m to x = 4.00 m. (Use any variable or symbol stated above as necessary.)
?U =

?K =

Explanation / Answer

a) F = -Ax + Bx^6, and F = -dU/dx, so

Fdx = -dU and

Fdx = -dU, so

U = -Fdx = (Ax - Bx^6)dx = (1/2)Ax^2 - (1/7)Bx^7 + c

If x =0, then U = c = 0, so

U = (1/2)Ax^2 - (1/7)Bx^7

b) U(4m) = (1/2)A(4 m)^2 - (1/7)B(4 m)^7 = (8*A - 2340.5*B) J

U(1.4 m) = (1/2)A(1.4 m)^2 - (1/7)B(1.4 m)^7 = (A - 1.5*B) J

U = (8*A - 2340.5*B) J - (A - 1.5*B) J = (7*A - 2339*B) J

Since the force is conserved, the energy is conserved as well (hence the name conserved), and

E1 = E2 --> U1 + K1 = U2 + K2, so

U = -K, and

K = (2339*B - 7*A) J

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