A. The speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring is 3.65m/s
B. The acceleration when the elevator is 1.00 {\rm m} below point where it first contacts a spring is 4m/s²
In calculating the speed of the elevator and acceleration, first we have to find the force of gravity F on the elevator, which is the force pulling the elevator in downward direction. Using the equation for force of gravity which is:
F = mg
Where:
Mass of the elevator; m= 2000kg
Acceleration due to gravity; g = 9.8m/s
2000kg × 9.8m/s²= 19600N
F = 19600
Force of opposing friction clamp of gravity = 17000N
Net force on the elevator = force of gravity - Force of opposing friction clamp
Net force on the elevator = 19600 - 17000
Net force on the elevator = 2600 N
We will also find the kinetic energy K.E; of the elevator at the point of contact with the spring using:
K.E = 1/2 mv²
Where
Mass of the elevator; m = 2000kg
Velocity of the elevator = 4.00m/s
K.E = (1/2)*2000kg*(4m/s)²
K.E = 16000J
The kinetic energy and energy gained will be absorbed by the spring across the next 2m
Therefore,
Energy; E = K.E + P.E
Where:
Kinetic energy K.E = 16000J
Potential Energy P.E = ?
P.E of spring = net force absorbed × distance at compression
Where:
Net force absorbed = 2600N
Distance at compression = 2.0m
P.E = 2600*2
P.E = 5200J
E = 16000J + 5200J
E = 21200J
Spring constant = k
To find k
Using:
E = (1/2)*k*(x)²
Where:
E = 21200J
k = ?
x = 2m
21200J = (1/2)*k*(2m)²
21200J*2 = (4m)k
K = 42400J/4m
K = 10600N/m
Therefore,
Acceleration at 1m compression = ?
Using:
F = K*X
Where
F is force provided by the spring = 10600N/m,
K = 10600 N/m
X = 1m
F = 10600N/m * 1m
F = 10600N (upward)
A. The speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?
Using:
Original Kinetic energy + net force on the elevator = final kinetic energy + spring energy
16000N + 2600N = (1/2)mv² + (1/2)k x²
18600 = (1/2)(2000)(v²) + (1/2)(10600N)(1²)
18600 = 1000(v²) + 5300
18600 - 5300 = 1000(v²)
13300 = 1000(v²)
V² = 13.300
V =3.65m/s
B. The acceleration of the elevator is 1.00m below point where it first contacts a spring
Spring constant = net force on the elevator + resultant force
Where:
Spring constant = 10600N
Net force on the elevator = 2600N
Resultant force = ?
10600N = 2600N + resultant force
Resultant force = 10600N - 2600N
Resultant force = 8000N
Using the equation for Newton's 2nd law where F = ma,
a = F/m
Where:
Resultant force; F =8000N
Mass of the elevator; m =2000kg)
a = 8000 / 2000
a = 4m/s²
Here's the complete question:
In a "worst-case" design scenario, a 2000kg elevator with broken cables is falling at 4.00m/s when it first contacts a cushioning spring at thebottom of the shaft. The spring is supposed to stop the elevator,compressing 2.00m as it does so. During the motion a safety clampapplies a constant 17000N frictional force to the elevator.
1. What is the speed of the elevator after it has moved downward 1.00m from the point where it first contacts aspring?
2. When the elevator is 1.00m below point where it first contacts a spring, what is its acceleration?
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