Question

A steel cable initially 7m long is used to strap three logs of timber onto a vehicle. The cable is under a tension of 400N. If the diameter of the cable is 4cm, i) By how much has it to be extended under this tension. ii) How much work has been done in producing this extension. (Young's modulus for steel is 2.0×10^11 N\m^2)​

Answer 1

i) The extension of the cable can be calculated using the formula:

ΔL/L = F/(A*Y)

where ΔL is the extension in length, L is the original length, F is the tension force, A is the cross-sectional area of the cable, and Y is the Young's modulus of steel.

First, we need to calculate the cross-sectional area of the cable:

A = πr^2 = π(0.02m)^2 = 0.00126 m^2

Now we can calculate the extension:

ΔL/L = 400N/(0.00126m^2 * 2.0×10^11 N/m^2) = 1.5873×10^-6

ΔL = (1.5873×10^-6)(7m) = 1.11×10^-5 m

So the cable has to be extended by 1.11×10^-5 m under this tension.

ii) The work done in producing this extension can be calculated using the formula:

W = (1/2)FΔL

where W is the work done, F is the tension force, and ΔL is the extension in length.

Substituting the given values:

W = (1/2)(400N)(1.11×10^-5 m) = 2.22×10^-3 J

Therefore, the work done in producing this extension is 2.22×10^-3 J.

Step-by-step explanation: