Question

# When a 75 g mass is suspended from a vertical spring, the spring isstretched from a length of 4.0 cm to a length of 7.0 cm. If themass is then pulled downward an additional 10 cm, what is the totalwork done against the spring force in joules?

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When a 75 g mass is suspended from a vertical spring, the spring isstretched from a length of 4.0 cm to a length of 7.0 cm. If themass is then pulled downward an additional 10 cm, what is the totalwork done against the spring force in joules?

2021-03-26
Given Data :- Intial length of the spring $$\displaystyle{\left({l}_{{{1}}}\right)}={4}{c}{m}={0.04}{m}$$.
Final length of the spring $$\displaystyle{\left({l}_{{{2}}}\right)}={7}{c}{m}.={0.07}{m}$$.
elongationof the wire $$\displaystyle{\left({e}\right)}={l}_{{{2}}}-{l}_{{{1}}}={0.07}-{0.04}={0.03}{m}$$.
mass attached to the spring (m) =75g.=0.075kg.
Acceleration due to gravity $$\displaystyle{\left({g}\right)}={9.8}\frac{{m}}{{s}^{{{2}}}}$$
Therefore Force constant $$\displaystyle{K}={\frac{{{m}{g}}}{{{e}}}}$$
$$\displaystyle{K}={\frac{{{0.075}\times{9.8}}}{{{0.03}}}}$$
$$\displaystyle\Rightarrow{K}={24.5}\frac{{N}}{{m}}$$
$$\displaystyle{W}={\frac{{{1}}}{{{2}}}}{K}{x}^{{{2}}}$$
Work Done in streched spring.
where x = streched length =10cm.= 0.1m.
$$\displaystyle{W}={\frac{{{1}}}{{{2}}}}\times{24.5}\times{\left({0.1}\right)}^{{{2}}}$$
$$\displaystyle\Rightarrow{W}={0.1225}{J}$$.