In a "worst-case" design scenario, a 2000 kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushioning spring at the bottom of the shaft. the spring is supposed to stop the elevator, compressing 2.00 m as it does so. spring coefficient is 10.6 kn/m. during the motion a safety clamp applies a constant 17000-n frictional force to the elevator.1. what is the speed of the elevator after it has moveddownward 1.00 {\rm m} from the point where it first contacts aspring? 2. when the elevator is 1.00 {\rm m} below point where it first contacts a spring, whatis its acceleration?

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

final velocity Vf = 3.65m/s

a = -4m/s

Explanation:

Vf = final velocity

Vi = initial velocity = 4m/s

m = 2000kg

F = 17000N

We know that elevator is stopped when spring is compressed by;

x = 2m

Step 1

Therefore;

½ kx^2 + F(x) - mgx = ½mv^2

= ½k(2^2) + (17000 × 2) - ( 2000 × 9.81 × 2) = ½(2000) (4^2)

= 2k + 34000 - 39240 = 16000

k = 10620N/m.

Speed when spring is compressed by x= 1

Wg + Wf + Waiting = ½m( vf^2 - vi^2)

2000(9.81)(1) - 17000(1) - ½(10620)(1^2) = ½(2000)(vf^2 - 4^2)

-2.69 = vf^2 - 16

Vf = 3.65m/s.

Step 2.

Net force when elevator spring is compressed x = 1

Fnet = mg - kx - FF

Fnet = 2000(9.81) - (10620 × 1) - 17000

Fnet = -8000N

Acceleration

a = F/m

a = -8000/2000

a = - 4m/s^2

im guessing the west of station b