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

Answer:

$\frac{cx^2}{2} -x^3 \Big|_0^4 = KE_f - KE_i$

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

$8c= -13 + 64$

$c = \frac{51}{8}= 6.375$

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:

$\frac{cx^2}{2} -x^3 \Big|_0^4 = KE_f - KE_i$

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

$8c= -13 + 64$

$c = \frac{51}{8}= 6.375$