Answer:
A. V =3.65m/s
B. a = 4m/s^2
Explanation:
Determine force of gravity (f) on the elevator.
f = mg
(m = 2000kg, g = 9.8m/s
2000kg × 9.8m/s^2= 19600N
Given,
Force of opposing friction clampforce of gravity = 17000N
the Net force on the elevator
= force of gravity - Force of opposing friction clamp
=19600 - 17000
= 2600 N
Lets determine the kinetic energy of the elevator at the point of contact with the spring
K.E = 1/2 m v^2
(m = 2000kg, v = 4.00m/s)
= (1/2) × 2000kg × (4m/s)^2
= 16000J
kinetic energy and energy gain will be absorbed by the spring across the next 2m
Therefore,
E = K.E + P.E
K.E = 16000J,
P.E of spring = net force absorbed × distance at compression
net force absorbed = 2600N and distance at compression = 2.0m)
P.E = 5200J
E = 16000J + 5200J
E = 21200 J
Note, spring constant wasn't given
Lets determine it's value
Using,
E = (1/2) × k × (x)^2
Where:
E = energy = 21200J, K = ?, X = 2m
21200J=(1/2) × k × (2m)^2
21200J × 2 =(4m)k
K = 42400J/4m
K = 10600 N/m
Therefore,
acceleration at 1m compression = ?
Using F = K × X
(F is force provided by the spring = 10600N/m, K = 10600 N/m and X = 1m)
= 10600N/m × 1m = 10600 N ( 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)mv^2 + (1/2)k x^2
18600 = (1/2)(2000)(v^2) + (1/2)(10600N)(1^2)
18600 = 1000(v^2) + 5300
18600 - 5300 = 1000(v^2)
13300 = 1000(v^2)
V^2 = 13.300
V =3.65m/s
The acceleration of the elevator is 1.00 {\rm m} below point where it first contacts a spring
Spring constant = net force on the elevator + resultant force
(Spring constant = 10600N, net force on the elevator = 2600N, resultant force = ?)
10600N = 2600N + resultant force
resultant force = 10600N - 2600N
=8000N
Therefore
F = ma
a = f/m
(a = ?, f =8000N and m =2000kg)
= 8000 / 2000
a = 4m/s^2
(It's accelerating upward, since acceleration is positive
