A 5-m-long steel cable is horizontally attached (just tautly) between two large trees. A 40-kg child hangs from the middle, and in doing so lowers the cable 6 cm from its original position. If the Young's modulus of the steel cable is $1 \times 10 ^ { 11 }$ Pa and the cross-sectional area is 0.7 cm $2$ , the distance that the steel cable is stretched is
A) 6 mm.
B) 6 m.
C) 6 cm.
D) 0.6 m.

Question

Quizplus · Accepted Answer

The correct answer is A because the distance that the steel cable is stretched is 6 mm. This can be calculated using the formula for the stress-strain relationship:$$\sigma = E \epsilon$$where $\sigma$ is the stress, $E$ is the Young's modulus, and $\epsilon$ is the strain. In this case, the stress is equal to the force divided by the cross-sectional area:$$\sigma = \frac{F}{A}$$where $F$ is the force and $A$ is the cross-sectional area. The strain is equal to the change in length divided by the original length:$$\epsilon = \frac{\Delta L}{L}$$where $\Delta L$ is the change in length and $L$ is the original length. Substituting these expressions into the stress-strain relationship, we get:$$\frac{F}{A} = E \frac{\Delta L}{L}$$Solving for $\Delta L$, we get:$$\Delta L = \frac{FL}{AE}$$Substituting the given values, we get:$$\Delta L = \frac{(40 \text{ kg})(9.8 \text{ m/s}^2)(5 \text{ m})}{(1 \times 10^{11} \text{ Pa})(0.7 \times 10^{-4} \text{ m}^2)} = 6 \times 10^{-3} \text{ m} = 6 \text{ mm}$$

A 5-M-Long Steel Cable Is Horizontally Attached (Just Tautly) Between 1×10111 \times 10 ^ { 11 }1×1011

A 5-M-Long Steel Cable Is Horizontally Attached (Just Tautly) Between $1 \times 10 ^ { 11 }$