You pull straight up on the string of a yo-yo with a force 0.19 N, and while your hand is moving up a distance 0.12 m, the yo-yo moves down a distance 0.30 m. The mass of the yo-yo is 0.058 kg, and it was initially moving downward with speed 3.4 m/s. (b) What is the new speed of the yo-yo

2.54 m/s

Explanation:

By conservation of energy,

work done by force due to hand = mechanical energy gained by yo-yo

So Fd = mgh + 1/2m(v₂² - v₁²)

where F = force on string = 0.19 N, d = distance moved by hand = 0.12 m, m = mass of yo-yo = 0.058 kg, g = acceleration due to gravity = 9.8 m/s², h = distance moved by yo-yo = 0.30 m, v₁ = initial speed of yo-yo = 3.4 m/s and v₂ = final speed of yo-yo = unknown

So,  Fd = mgh + 1/2m(v₂² - v₁²)

Fd/m = gh + 1/2(v₂² - v₁²)

Fd/m - gh = 1/2(v₂² - v₁²)

2(Fd/m - gh) = (v₂² - v₁²)

(v₂² - v₁²) = 2(Fd/m - gh)

v₂²  = v₁² + 2(Fd/m - gh)

taking square-root of both sides, we have

v₂  = √[v₁² + 2(Fd/m - gh)]

Substituting the values of the variables into the equation, we have

v₂  = √[(3.4 m/s)² + 2(0.19 N × 0.12 m/0.058 kg - 9.8 m/s² × 0.30 m)]

v₂  = √[(3.4 m/s)² + 2(0.19 N × 0.12 m/0.058 kg - 9.8 m/s² × 0.30 m)]

v₂  = √[(3.4 m/s)² + 2(0.0228 Nm/0.058 kg - 9.8 m/s² × 0.30 m)]

v₂  = √[(3.4 m/s)² + 2(0.393 Nm/kg - 2.94 m²/s²)]

v₂  = √[(3.4 m/s)² + 2(-2.547 m²/s²)]

v₂  = √[11.56 m²/s² - 5.094 m²/s²)]

v₂  = √[6.466 m²/s²]

v₂ = 2.54 m/s

this allows people to make the best decision or pursue the best option. when alternatives are weighed using their advantages and disadvantages, the option whose disadvantages outweigh the advantages (based on a qualitative or quantitative analysis) are shunned while those whose advantages outweigh their disadvantages are pursued.