Answer:
a) [tex]v \approx 2\,\frac{m}{s}[/tex], b) [tex]v \approx 2\,\frac{m}{s}[/tex]
Explanation:
a) The package handling system is modelled by using the Principle of Energy Conservation and the Work-Energy Theorem:
[tex]U_{k} = K_{t} + W_{loss}[/tex]
The final kinetic energy is:
[tex]K_{t} = U_{k} - W_{loss}[/tex]
[tex]\frac{1}{2}\cdot m \cdot v^{2} = \frac{1}{2}\cdot k \cdot x^{2} + \mu_{k}\cdot m\cdot g\cdot x[/tex]
[tex]v^{2} = \frac{k}{m} \cdot x^{2} + 2\cdot \mu_{k}\cdot g\cdot x[/tex]
The final speed is:
[tex]v = \sqrt{\frac{k}{m}\cdot x^{2}+2\cdot \mu_{k}\cdot g \cdot x }[/tex]
[tex]v = \sqrt{\frac{48\,\frac{N}{m} }{1.60\,kg}\cdot (0.280\,m)^{2}+2\cdot (0.3)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.280\,m) }[/tex]
[tex]v \approx 2\,\frac{m}{s}[/tex]
b) The maximum speed of the box during its motion occurs at the beginning of the movement, since the work done by friction is non-conservative and reduces speed as box travels on the horizontal surface. Hence, the maximum velocity is:
[tex]v \approx 2\,\frac{m}{s}[/tex]
