Answer:
W = 1.875 J
Explanation:
For this exercise let's use the relationship between work and kinetic energy
W = ΔK
The kinetic energy of rotational motion is
K₀ = ½ I w²
we can assume that the box is small, so it can be treated as a point object, with moment of inertia
I = m rₐ²
angular and linear velocity are related
v = w r
w = v / r
we substitute in the equation, for point A
K₀ = ½ (m rₐ²) (v / rₐ)²
K₀ = ½ m v²
For the final point B, as the system is isolated the angular momentum is conserved
initial L₀ = Io wo
final L_f = I_f w_f
L₀ = L_f
I₀ w₀ = I_f w_f
(m rₐ²) w₀ = (m [tex]r_{b} ^2[/tex]) w_f
w_f = (rₐ/r_b)² w₀
with this value we find the final kinetic energy
K_f = ½ I_f w_f²
K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)² w₀) ²
K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]
we substitute in the realcion of work
W = K_f - K₀
W = ½ m [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²
W = ½ m [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²
W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2
W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]
let's calculate
W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)
W = 0.625 (3)
W = 1.875 J
