Question

acts on a particle as the particle moves along an x axis, with in newtons, x in meters, and c a constant. At x = 0 m, the particle's kinetic energy is 22.0 J; at x = 4.00 m, it is 9.00 J. Find c.

Answer 1

`(cx^2)/(2) -x^3 \Big|_0^4 = KE_f - KE_i`

`8c - 64 = 9 - 22 J = -13 J`

`8c= -13 + 64`

`c = (51)/(8)= 6.375`

Step-by-step explanation:

Assuming the following question: A force F = (cx 3.00x2 li 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 22.0 J; at x = 4.00 m, it is 9.00 J. Find C.

Solution to the problem

We know that the work for this case since we don't have change of potential energy is given by:

`W = KE_f -KE_i`

Where KE means kinetic energy

`KE_f = 9 J, KE_i = 22 J`

And we know that the work can be expressed like this:

`W = \int_0^4 (cx - 3x^2) dx = KE_f - KE_i`

If we integrate the left part we got:

`(cx^2)/(2) -x^3 \Big|_0^4 = KE_f - KE_i`

`8c - 64 = 9 - 22 J = -13 J`

`8c= -13 + 64`

`c = (51)/(8)= 6.375`