A single conservative force acting on a particle varies as = (-Ax + Bx4) N, wher

Question

A single conservative force acting on a particle varies as = (-Ax + Bx4) 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 = 3.80 m. (Use any variable or symbol stated above as necessary.)
?U =

?K =

Explanation / Answer

Since F = -grad(U), then U = -line integral[a,b]() where F and dx are vectors and "" is the scalar product, a and b are limits So you need to do an indefintie integral --> U = - line integral() U = - line integral([-Ax + Bx^4] dx) = +A/2 x^2 - B/3 x^5 + k where k is a constant U(0) = 0 --> k = 0 U(x) = A/2 x^2 - B/5 x^5 Part b. Since F is conservative, change in potential energy is just the difference between the potential at the two points in the field. So DU = U(1.4) - U(3.4) = -4.8 A +89.79 B Since ther eis a change in potential energy there has to be a corresponding change in kinetic energy.

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