Question

10 If a force of 80N extends a spring of natural length 8m by 0.4m what will be the length of the spring when the applied force is 100N
0.3m
0.5m
1.0m
2.0m

11 A wire of cross-sectional area of
6×10−5m2
and length 50cm stretches by 0.2mm under a load of 3000N. Calculate the Young’s modulus for the wire
8×1010Nm−2
1.25×1011Nm−2
2.5×1011Nm−2
5×1011Nm−2

Answer 1

Answer on Question #58117, Physics / Mechanics | Relativity |

10 If a force of 80N extends a spring of natural length 8m by 0.4m what will be the length of the spring when the applied force is 100N

0.3m

0.5m

1.0m

2.0m

Solution:

First of all we need to find the spring constant (from Hooke's law):

$$F _ {1} = k x _ {1}$$

$F_{1}$ – applied force in the first case, $x_{1}$ – extension.

$$k = \frac {F _ {1}}{x _ {1}} = \frac {8 0 \mathrm {N}}{0 . 4 \mathrm {m}} = 2 0 0 \frac {\mathrm {N}}{\mathrm {m}}$$

So, the length of the spring in the second case will be greater its natural length:

$$x _ {2} = \frac {F _ {2}}{k} = \frac {1 0 0 \mathrm {N}}{2 0 0 \mathrm {N / m}} = 0. 5 \mathrm {m}$$

And the total length in second case:

$$l = l _ {0} + x _ {2} = 8 + 0. 5 = 8. 5 \mathrm {m}$$

Answer: extension = 0.5 m;

total length when the applied force is $100\mathrm{N} = 8.5\mathrm{m}$

11 A wire of cross-sectional area of $6 \times 10^{-5} \mathrm{~m}^2$ and length 50cm stretches by 0.2mm under a load of 3000N. Calculate the Young's modulus for the wire

$8 \times 10^{10} \mathrm{Nm}^{-2}$

$1.25 \times 10^{11} \mathrm{Nm}^{-2}$

$2.5 \times 10^{11} \mathrm{Nm}^{-2}$

$5 \times 10^{11} \mathrm{Nm}^{-2}$

**Solution:**

Young's modulus $E$, is

$$E = \frac{F / A}{\Delta L / L}$$

where

- E is the Young's modulus (modulus of elasticity)

- F is the force exerted on an object under tension;

- A is the actual cross-sectional area through which the force is applied;

- $\Delta L$ is the amount by which the length of the object changes;

- L is the original length of the object.

Hence,

$$E = \frac{3000 / 6 \cdot 10^{-5}}{0.2 \cdot 10^{-3} / 0.5} = 1.25 \cdot 10^{11} \mathrm{Nm}^{-2}$$

**Answer:** $1.25 \cdot 10^{11} \mathrm{Nm}^{-2}$

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