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

layers of the earth affect the waves and these effects are seen on instruiments such as seimometers