Question

A force F→=(cx-3.00x2)iˆ 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 20.0 J; at x = 2.00 m, it is 11.0 J. Find c.

Answer 1

Step-by-step explanation:

F = cx - 3x²

At x = 0 , Kinetic energy = 20 J

at x = 2 m, the kinetic energy = 11 J

Use work energy theorem

`W = \int_(0)^(2)F(x) dx = \Delta K`

`W = \int_(0)^(2)(cx-3x^(2)) dx = 11-20`

`(cx^(2))/(2)-x^(3)=-9`

2c - 8 = - 9

2c = - 1

c = - 0.5

Answer 2

Step-by-step explanation:

Work done = ∫Fdx

= ∫(cx-3.00x²) dx

[ c x² / 2 - 3 x³ / 3 ]₀²

= change in kinetic energy

= 11-20

= - 9 J

[ c x² / 2 - x³ ]₀² = - 9

c x 2² / 2 - 2³ = -9

2c - 8 = -9

2c = -1

c = - 1/2