A force F=(cx-3.00x2)i acts on a particle as the particle moves along an x axis, with F in newtons, x in meters, and c a constant. At x = 0 m, the particle's kinetic energy is 18 J; at x = 5 m, it is 12 J. Find c.

The value of c is 9.52

Explanation:

Given;

force F = (cx-3x²)i which act on the particle through a distance x

The kinetic of the particle = Fdx

given Kinetic energy as 18 J

KE = ∫Fdx

KE = ∫(cx-3x²)dx

KE = ¹/₂cx² - ¹/₃(3x³) + k

KE =  ¹/₂cx² - x³ + k

where, k is the constant of integration

Given at x = 0 m, KE = 18 J, substitute in this in the integral equation

18 =  ¹/₂c (0)² - (0)³ + k

k = 18

Also, Given at x = 5 m, KE = 12 J substitute in this as well in the integral equation;

12 = ¹/₂c(5)² - (5)³ + 18

12 = 12.5c - 125 + 18

12.5c = 12 + 125 - 18

12.5c = 119

c = 119/12.5

c = 9.52

gets less

6 cos 34, 6 sin 34 trig