If a nucleus captures a stray neutron, it must bring the neutron to a
stop within the diameter of the nucleus by means of the strong force.
That force, which "glues" the nucleus together, is essentially zero
outside the nucleus. Suppose that a stray neutron with an initial
speed of 1.4 x 10^7 m/s is just barely captured by a nucleus with
diameter d = 1.0 x 10^-14 m. Assuming that the force on the neutron is
constant, find the magnitude of that force. The neutron's mass is 1.67
x 10^-27kg.
 
Given:
V(final)^2  - V(initial)^2 = 2(a)(delta x)  from this
calculate the acceleration assuming that (delta x) = diameter of the
nucleus ; then apply  the 2nd law in the form:
F = (m)(a) where (m) = mass of neutron.
Remember that the neutron is being brought to a stop, so V(final) is
zero. Solve for F and give the answer in Newtons, which is kg*m/s^2.

Hi!!

From the equation V(final)^2  - V(initial)^2 = 2*(a)*(delta x) we have
that (a) is equal to:

(a) = [V(final)^2  - V(initial)^2] / [2*(delta x)] =
  
   = [0^2 - (1.4*10^7 m/s)^2] / [2*(1.0*10^-14 m)] =

   = [0 - 1.96*10^14 (m/s)^2] / [2.0*10^-14 m] =

   = [-1.96*10^14 (m/s)^2] / [2.0*10^-14 m] =

   = -9.8*10^27 m/s^2


Now we can find F.

F = (m)*(a) = 
  
  = 1.67*10^-27 kg * 9.8*10^27 m/s^2 =  
 
  = 16.366 kg*m/s^2 =

  = 16.366 Newtons

The magnitude of the force is 16.366 N .


Hope that this helps you. Feel free to request for a clarification if you need it.

Regards.
livioflores-ga