Suppose a force of 30 N is required to stretch and hold a spring 0.1 m from its equilibrium position. a. Assuming the spring obeys Hooke's law, find the spring constant k. b. How much work is required to compress the spring 0.3 m from its equilibrium position? c. How much work is required to stretch the spring 0.2 m from its equilibrium position? d. How much additional work is required to stretch the spring 0.1 m if it has already been stretched 0.1 m from its

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xero099 xero099 · Answer 1 · 2020-03-11T06:08:01+01:00

Answer:

a) [tex]k = 300\,\frac{N}{m}[/tex], b) [tex]\Delta U_{k} = 13.5\,J[/tex], c) [tex]\Delta U_{k} = 6\,J[/tex], d) [tex]\Delta U_{k} = 4.5\,J[/tex]

Step-by-step explanation:

a) The spring constant is calculated by using this expression:

[tex]k = \frac{F}{x}[/tex]

[tex]k = \frac{30\,N}{0.1\,m}[/tex]

[tex]k = 300\,\frac{N}{m}[/tex]

b) The work needed to compress the spring from its initial position is:

[tex]\Delta U_{k} = \frac{1}{2}\cdot k \cdot (x_{f}^{2}-x_{o}^{2})[/tex]

[tex]\Delta U_{k} = \frac{1}{2}\cdot (300\,\frac{N}{m} )\cdot [(-0.3\,m)^{2}-(0\,m)^{2}][/tex]

[tex]\Delta U_{k} = 13.5\,J[/tex]

c) The work needed to stretch the spring is:

[tex]\Delta U_{k} = \frac{1}{2}\cdot (300\,\frac{N}{m} )\cdot [(0.2\,m)^{2}-(0\,m)^{2}][/tex]

[tex]\Delta U_{k} = 6\,J[/tex]

d) The work need to stretch the spring is:

[tex]\Delta U_{k} = \frac{1}{2}\cdot (300\,\frac{N}{m} )\cdot [(0.2\,m)^{2}-(0.1\,m)^{2}][/tex]

[tex]\Delta U_{k} = 4.5\,J[/tex]

nwandukelechi nwandukelechi · Answer 2 · 2020-03-11T09:29:47+01:00

Step-by-step explanation:

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