Question

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

Answer 1

Answer:

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

Step-by-step explanation:

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

$k = \frac{F}{x}$

$k = \frac{30\,N}{0.1\,m}$

$k = 300\,\frac{N}{m}$

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

$\Delta U_{k} = \frac{1}{2}\cdot k \cdot (x_{f}^{2}-x_{o}^{2})$

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

$\Delta U_{k} = 13.5\,J$

c) The work needed to stretch the spring is:

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

$\Delta U_{k} = 6\,J$

d) The work need to stretch the spring is:

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

$\Delta U_{k} = 4.5\,J$

Answer 2

Step-by-step explanation:

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