Answer:
a) [tex]s = 1.534\,m[/tex], b) [tex]s = 6.135\,m[/tex]
Explanation:
a) The energy equation for the crate is modelled after the Principle of Energy Conservation and Work-Energy Theorem. Changes in gravitational potential energy can be neglected due to the information of a horizontal surface:
[tex]K_{A} = W_{loss}[/tex]
[tex]\frac{1}{2}\cdot m \cdot v^{2} = \mu_{k} \cdot m \cdot g\cdot s[/tex]
The distance that crate needs to cover before stopping is:
[tex]s = \frac{v^{2}}{2\cdot \mu_{k}\cdot g}[/tex]
[tex]s = \frac{(1.90\,\frac{m}{s} )^{2}}{2\cdot (0.120)\cdot (9.807\,\frac{m}{s^{2}} )}[/tex]
[tex]s = 1.534\,m[/tex]
b) The stopping distance is:
[tex]s = \frac{(3.80\,\frac{m}{s} )^{2}}{2\cdot (0.120)\cdot (9.807\,\frac{m}{s^{2}} )}[/tex]
[tex]s = 6.135\,m[/tex]