A 3.14-m long steel cable with a cross-sectional area of 1.42 cm² is attached to a 894-kg car ( )that has fallen into a pond ( ).How much will the cable stretch as it pulls the car upward at a constant speed? Assume the car is pulled slowly enough to ignore the effects of resistance and ignore the weight of the steel cable.(Young's modulus for steel is 200 x 10⁹ N/m².)

Question

Quizplus · Accepted Answer

First, we need to find the tension in the cable, which is equal to the weight of the car:
T = mg = (894 kg)(9.8 m/s^2) = 8761.2 N
Next, we can use Hooke's Law (F = kx) to find the amount of stretch in the cable:
T/A = YΔL/L
where T is the tension, A is the cross-sectional area, Y is the Young's modulus, ΔL is the change in length, and L is the original length.
Solving for ΔL, we get:
ΔL = TL/(AY)
Plugging in the given values, we get:
ΔL = (8761.2 N)(3.14 m)/(1.42 cm^2)(200 × 10^9 N/m^2) = 0.85 mm
Therefore, the cable will stretch by 0.85 mm as it pulls the car upward at a constant speed. The answer is D.

A 314-M Long Steel Cable with a Cross-Sectional Area of 1