An elevator manufacturing company is stress-testing a new elevator in an airless test shaft. The elevator is traveling at an unknown velocity when the cable snaps. The elevator falls 2.50 meters before hitting the bottom of the shaft. The elevator was in free fall for 0.900 seconds. Determine its velocity when the cable snapped. As usual, up is the positive direction.
The average velocity during the time is
2.50/.9=(Vf+Vi)/2
or Vf+Vi=5.56m/s
Vf=Vi+gt=Vi+8.82
Vi=Vf-8.82=-Vf+5.56
2Vf=8.82+5.56 solve for Vf
thank you
8.82
To determine the velocity of the elevator when the cable snapped, we can use the kinematic equation for motion during free fall:
Δy = v₀t + (1/2)gt²
where:
Δy is the displacement (2.50 meters, negative because it is downward)
v₀ is the initial velocity (which we want to find)
t is the time (0.900 seconds)
g is the acceleration due to gravity (-9.8 m/s², negative because it is downward)
Rearranging the equation, we get:
v₀ = (Δy - (1/2)gt²) / t
Substituting the values, we have:
v₀ = (2.50 - (1/2)(-9.8)(0.900)²) / 0.900
Simplifying the equation, we get:
v₀ = (2.50 + 3.9324) / 0.900
v₀ = 6.4324 / 0.900
v₀ ≈ 7.14 m/s
Therefore, the velocity of the elevator when the cable snapped was approximately 7.14 m/s downward.