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algol [13]
3 years ago
10

Answer both questions please! Due today.

Mathematics
1 answer:
podryga [215]3 years ago
8 0
The answer for the problem is B
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A. Work of 5 Joules is done in stretching a spring from its natural length to 16 cm beyond its natural length. What is the force
kirza4 [7]

Answer:

(a) 62.5 N

(b) 694.4 N/m

(c) 41.7 N

Step-by-step explanation:

Work done in stretching a spring is

W = \frac{1}{2}Fe=\frac{1}{2}ke^2 since F=ke

F is the applied force, E is the elongation and k is the spring constant.

(a) Here, e = 16 cm = 0.16 m

W = \frac{1}{2}Fe

F = \dfrac{2W}{e} = \dfrac{2\times5}{0.16}=62.5\text{ N}

(b) Here, e = 12 cm = 0.12 m

W =\frac{1}{2}ke^2

k=\dfrac{2W}{e^2}=\dfrac{2\times5}{0.12^2}=694.4\text{ N/m}

(c) The restoring force is the same as the applied force.

F = \dfrac{2W}{e} = \dfrac{2\times5}{0.12}=41.7\text{ N}

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