Question

O 444/6.66 points Previous Answers SerCP10 9.P016 A high-speed lifting mechanism supports a(n) 820-kg object with a steel cable that is 40.0 m long and 4.00 cm in cross-sectional area. (a) Determine the elongation of the cable. (Enter your answer to at least two decimal places.) 4.018 mm (b) By what additional amount does the cable increase in length if the object is accelerated upwards at a rate of 3.5 m/s ? X Enter a number. differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. mm (c) what is the greatest mass that can be accelerated upward at 3.5 m/s2 if the stress in the cable is not to exceed the elastic limit of the cable, which is 2.2 6616 5 kg Need Help? Read I Practice Another Version Submit Answer Save Progress SerCP10 9 P 053 M FB 1 Previous Anowers 6.76-6.76 points 15. a x N hp 144 4- & A

Answer 1

As you've already solved part a and c if you need help with any of those parts let me know.

Part (b)

From, Young’s modulus:
Y = Stress/Strain
Stress = Y*strain
Strain = dL/L0
Stress = Force/Area
Y = F*L0/(A*dL)

A = pi*r^2

So, length of stretched wire = dL = F*L0/(A*Y)

L0 = 40.0 m

A = 4.00 cm^2 = 4.00*10^-4 m^2

Y = 2*10^11 Pa

Now when object is accelerating in upward direction, then Using Newton's 2nd law:

Fnet = m*a

(Here F force is in upward direction and weight is in downward direction, So)

F - W = m*a

F = W + m*a

W = Weight of object = m*g

F = m*(g + a)

m = 820 kg & a = 3.5 m/sec^2

Using above values:

dL = (820*(9.8 + 3.5)*40.0)/(4.00*10^-4*2*10^11)

dL = 0.005453 m = 5.453*10^-3 m = 5.453 mm

Now additional increase in length will be:

from part A length is increased by 4.018 mm, So

additional increase = 5.453 - 4.018 = 1.435 mm

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