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

Answer:

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

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

$W=\frac{1}{2} mv^{2} -0$

The velocity is:

$v=\sqrt{\frac{2W}{m} } =\sqrt{\frac{2*258352.5}{925} } =23.63m/s$

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

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

Replacing:

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

Answer 2

Answer:

a) = 258352.5J

b) = 23.63 m/s

c) = 1.8m

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