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

Answer:

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

Answer 2

Answer:

Explanation:

F = cx - 3x²

At x = 0 , Kinetic energy = 20 J

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

Use work energy theorem

$W = \int_{0}^{2}F(x) dx = \Delta K$

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

$\frac{cx^{2}}{2}-x^{3}=-9$

2c - 8 = - 9

2c = - 1

c = - 0.5