Question

An elevator cable breaks when a 925-kg elevator is 28.5 m above the top of a huge spring Ak = 8.00 * 104 N????mB at the bottom of the shaft. Calculate (a) the work done by gravity on the elevator before it hits the spring; (b) the speed of the elevator just before striking the spring; (c) the amount the spring compresses (note that here work is done by both the spring and gravity).

Answer 1

a) = 258352.5J

b) = 23.63 m/s

c) = 1.8m

Step-by-step explanation:

Data;

Mass = 925kg

Distance (s) = 28.5m

Force constant (k) = 8.0*10⁴ N/m

g = 9.8 m/s²

a) = work = force * distance

But force = mass * acceleration

Force = 925 * 9.8 = 9065N

Work = F * s = 9065 * 28.5 = 258352.5J

b) acceleration (a) = (v² - u²) / 2s

a = v² / 2s

v² = a * 2s

v² = 9.8 * (2 * 28.5)

v² = 9.8 * 57

v² = 558.6

v = √(558.6)

V = 23.63 m/s

C). The work stops when the work done to raise the spring equals the work done to stop it by the spring

W = ½kx²

258352.5 = ½ * 8.0*10⁴ * x²

(2 * 258352.5) = 8.0*10⁴x²

516705 = 8.0*10⁴x²

X² = 516705 / 8.0*10⁴

X² = 6.46

X = √(6.46)

X = 2.54m

The compression was about 2.54m

Answer 2

a) The work done by gravity on the elevator is 258352.5 J

b) The speed of the elevator just before striking the spring is 23.63 m/s

c) The amount the spring compresses is 2.7 m

Step-by-step explanation:

a) The work is equal to:

`W=Fdcos\theta \\W=mgdcos\theta`

Where

m = 925 kg

g = 9.8 m/s²

d = 28.5 m

θ = 0°

Replacing:

`W=925*9.8*28.5*cos0=258352.5J`

b) The work done by gravity on the elevator is equal to the change of kinetic energy:

W = ΔEk

`W=(1)/(2) mv^(2) -0`

The velocity is:

`v=\sqrt{(2W)/(m) } =\sqrt{(2*258352.5)/(925) } =23.63m/s`

c) The total work is equal to the sum to of the change of kinetic energy and the spring:

`W_(g) +W_(spring) =0\\(k)/(2) x^(2) -(mg)x-mgd=0\\x=\frac{mg+-\sqrt{m^(2) g^(2)+2kmgd } }{k}`

Replacing:

`x=\frac{925*9.8+-\sqrt{925^(2)*9.8^(2)+(2*8x10^(4)*925*9.8*28.5) } }{8x10^(4) } =2.7m`