Physics: By how much does a 65.0-kg mountain climber stretch her 0.800-cm diameter nylon rope when she hangs 35.0 m below a rock outcropping? Young s modulus for the nylon rope is 5×109N/m2.

By how much does a 65.0-kg mountain climber stretch

pandadude123

19.01.2022

Explanation:

Expression for Young's modulus is represented as follows.

Y = $\frac{\text{shear stress}}{\text{shear strain}}$

= $\frac{\frac{F}{A}}{\frac{\Delta L}{L_{o}}}$

where, Y = Young's modulus

F = force which acts perpendicular to the surface

A = area of the surface

$L_{o}$ = original length of the rope

$\Delta L$ = elongation in the rope

The given data is as follows.

r = 0.008 m, mass (m) = 65 kg, g = 9.8 m/s

S = $5 \times 10^{9} N/m^{2}$

$L_{o}$ = 35 m

Force perpendicular to the surface is calculated as follows.

F = mg

= $65 kg \times 9.8 m/s$

= 637 N

and, Area = $\pi r^{2}$

= $3.14 \times (0.008)^{2}$

= $20.096 \times 10^{-5}$

Putting the given values into the above formula as follows.

$\Delta L = \frac{1}{Y} \times \frac{F}{A} \times L_{o}$

= $\frac{1}{5 \times 10^{9}} \times \frac{637 N}{20.096 \times 10^{-5}} \times 35 m$

= $221.879 \times 10^{-4}$

= $2.21 \times 10^{-2} m$

Thus, we can conclude that by $2.21 \times 10^{-2} m$ does a 65.0 kg mountain climber stretch her 0.800 cm diameter nylon rope.

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