Think I figured out part of the problem?

Question

A single conservative force F(x) = bx + a acts on a 3.28 kg particle, where x is in meters, b = 6.49 N/m and
a = 4.05 N.

#1 As the particle moves along the x axis from X1 = 0.936 m to X2 = 4.03 m, calculate the work done by thsi force in units of J.

#2 Calculate the change in potential energy of the particle in units of J.
THINK THE ANSWER TO THIS IS 62.3895 ??

#3 Calculate the particle's initial kenetic energy at X1 if its finals speed at X2 is 19.8 m/s (in units of J).

Answer 1

1) W = int((F∙dx) = b*int(x,dx) + a*int(dx)
= b(1/2)x^2 + a*x
Now apply your limits and plug in your constants:
W = (6.49)*(1/2)*[4.03^2 - 0.936^2] + (4.05)*[4.03 - 0.936 ]
= 49.86 + 12.53 = 62.39 J

2) F is a conservative force, so the change in potential energy is equal to NEGATIVE the work done (because the force did work ON the particle, the particle must have a lower potential energy...like how gravity does work on an apple to make it fall):
[delta]U = -W = -62.39 J

3) This may look like a basic kinematics problem...but beware, its not. Because the force (and consequently the acceleration) is non-constant, you CANNOT use the basic kinematic equations.
Luckily the alternative method is just as easy.
A conservative force is one which causes no energy loss (i.e. energy is "conserved"), so you can apply the conservation of energy:
Ei = Ef ───> Ui + Ki = Uf + Kf
Solve for Ki:
Ki = Kf + (Uf - Ui)
= (1/2)*m*vf² - W (remember, the change in pot. energy is just W)
= (1/2)*(3.28)*(19.8)² - (62.39) = 580.6 J

Answer 2

I did that real quick in my head and you are off by .0001

Answer 3

sounds good

Answer 4

WOW! do you do to yale???????