Work and Kinetic Energy problem

[maniacp08]

A force Fx acts on a particle that has a mass of 1.5kg. The force is related to the position x of the particle by the formula Fx = Cx^3, where C = .50 if x is in meters and Fx is in Newtons.

a) What are the SI units of C?
b)Find the work done by this force as the particle moves from x=3.0m to x=1.5m
c)At x = 3.0m, the force points opposite the direction of particle's velocity(speed is 12.0m/s). What is its speed at x=1.5m?

For A:
Isn't C just a constant? What units is this in?

For B:
I have to integrate Fx and the integral of F(x)dx = x^4/4
Since C is a constant I can put it in front of the integral so is 1/4 C x^4
When I integrate shouldn't there be a + K? 1/4 C x^4 + K
X1 = 3.0 and X2 = 1.5
Plug in and solve?

For C:
Using answer from Part B = Work total
Work total = 1/2 M Vf^2 - 1/2 M Vi^2
where Vi = 12m/s?

 

[Doc Al]

For A:
Isn't C just a constant? What units is this in?

Just because C is a constant doesn't mean it has no units. Hint: For any equation, the units on each side must match.

For B:
I have to integrate Fx and the integral of F(x)dx = x^4/4
Since C is a constant I can put it in front of the integral so is 1/4 C x^4

Good.

When I integrate shouldn't there be a + K? 1/4 C x^4 + K
X1 = 3.0 and X2 = 1.5
Plug in and solve?

Since you are doing a definite integral (between limits) the integration constant goes away. Just evaluate between those limits.

For C:
Using answer from Part B = Work total
Work total = 1/2 M Vf^2 - 1/2 M Vi^2
where Vi = 12m/s?

Good. Use the work-energy theorem.