Answer:
a) U = 2 (x²) - 3 (x³)
b) U = - 170,839 J
c) ΔK = 341,678 J
d) W = 170,839 J
Explanation:
Force and potential energy are related by the equation
F = - dU / dx
a) Let's look for potential energy
dU = - F dx
We integrate
dU = - ∫ (-4x +9 x²) dx
U- Uo = (4 x² / 2 - 9 x³ / 3)
Where they indicate that U₀ = 0 for x = 0.
U = 2 (x²) - 3 (x³)
b) We integrate and evaluate for the points of interest. Let's evaluate between the lower limit x₁ = 1.20 m and the upper boundary x₂ = 4.10 m
U = 2 (4.1² - 1.2²) - 3 (4.1³ -1.2³)
U = 30.74 - 201.579
U = - 170,839 J
d) work is defined by
W = ∫ F. dx
Since the force is in the x direction the scale product is reduced to the algebraic product
W = ∫ (-4 x + 9 x²) dx
W = - 2 x² + 3 x³
We evaluate between the limits
W = -2 (4.1² -1.2²) + 3 (4.1³- 1.2³)
W = -30.74 + 201.579
W = 170,839 J
c) let's look for the kinetic energy of the particle
W = ΔK + ΔU
ΔK = W - ΔU
ΔK = 170,839 - (-170,839)
ΔK = 341,678 J
Answer:
(a) ux = 2x² -3x³
(b) Change in potential energy = -170.839j
(c) Change in kinetic energy = 170.839j
(d) Workdone by force = 170.839j
Explanation:
See the attached for the calculation
