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.

 

[thomate1]

A) The SI units of C would be Newton-meters per meter cubed (N·m/m^3), which simplifies to N/m^2 or Pascals (Pa). This can also be written as Joules per meter cubed (J/m^3).

B) To find the work done by this force, we can use the work-energy theorem which states that the work done by a force is equal to the change in kinetic energy of the particle. In this case, the force is varying with position, so we need to use the work-energy theorem in integral form: W = ∫F(x)dx. Plugging in the given force formula, we get W = ∫0.5x^3dx. Evaluating this integral from x=3.0m to x=1.5m, we get W = 0.5(1.5^4 - 3.0^4) = -20.25 J. Note that the negative sign indicates that the work is done in the opposite direction of the force, as the particle is moving from x=3.0m to x=1.5m.

C) To find the speed of the particle at x=1.5m, we can use the conservation of energy principle. The total energy of the particle is equal to the work done by the force plus its initial kinetic energy: E = W + 0.5mv^2. Plugging in the values from Part B and the given initial speed of 12.0 m/s, we get E = -20.25 J + 0.5(1.5)(12.0)^2 = 81.75 J. To solve for the final speed, we can rearrange the equation to v = √(2(E-W)/m). Plugging in the values, we get v = √(2(81.75 - (-20.25))/1.5) = 12.26 m/s. Therefore, the speed of the particle at x=1.5m is 12.26 m/s.