Question

a 200 kg load is hung on a wire of length of 4.00 m, cross sectional area 0.200 x 10^-4 m ^2 ,and Young's modulus 8.00 x 10^10 N/m^2.What is its increase in length?

Answer 1

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Answer on Question #37964, Physics, Mechanics

Question:

a 200 kg load is hung on a wire of length of 4.00 m, cross sectional area 0.200 x 10^4 m^2, and Young's modulus 8.00 x 10^10 N/m^2. What is its increase in length?

Answer:

Hooke's law can be expressed in equation form as follows:

$$\frac {F}{A} = E \frac {\Delta l}{l}$$

where $E$ is Young's modulus, $A$ - cross section area, $l$ - length of the wire, $F$ - force, $\Delta l$ is increase in length.

In our case force equals weight: $F = mg$

Therefore, stretch of the wire equals:

$$\Delta l = \frac {mg}{A} \frac {l}{E} \cong 4.91 \, \text{mm}$$

Answer: 4.91 mm