Question

A force F=(cx-3.00x2)i acts on a partical as theparticle

moves along an x axis, with F in newtons, x in meters andc a
constant..At x=0, the paricle's kinetic energy is 20.0 J;atx=3.00m
it is 11.0 J......FIND c...

Answer 1

The value of 'c' is 4.

Step-by-step explanation:

We know from the basic relation between work and force is

`W_(2)-W_(1)=\int_{x_(1)}^{x_(2)}F\cdot dx`

Now since the energy of the particle changes from 20 Joules to 11 Joules as position changes from x=0 to x=3.00 thus applying the given vales in above relation we get

`11-20=\int_(0)^(3)(cx-3x^(2))\cdot dx\\\\-9=[(cx^2)/(2)]_0^3-[(3x^3)/(3)]_0^3\\\\-9=(9c)/(2)-27\\\\4.5c=18\\\\\therefore c=4`