Answer:
[tex]v\approx 8.570\,\frac{m}{s}[/tex]
Explanation:
The equation of equlibrium for the box is:
[tex]\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a[/tex]
The formula for the acceleration, given in [tex]\frac{m}{s^{2}}[/tex], is:
[tex]a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}[/tex]
Velocity can be derived from the following definition of acceleration:
[tex]a = v\cdot \frac{dv}{dx}[/tex]
[tex]v\, dv = a\, dx[/tex]
[tex]\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx[/tex]
[tex]\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx[/tex]
[tex]\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}[/tex]
[tex]v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }[/tex]
The speed after the box has travelled 17 meters is:
[tex]v\approx 8.570\,\frac{m}{s}[/tex]
