Answer:
The distance travel before stopping is 1.84 m
Explanation:
Given :
coefficient of kinetic friction [tex]\mu_{k} = 0.250[/tex]
Zak's speed [tex]v = 3 \frac{m}{s}[/tex]
Gravitational acceleration [tex]g = 9.8[/tex] [tex]\frac{m}{s^{2} }[/tex]
Work done by frictional force is given by,
[tex]W = \Delta K[/tex]
[tex]\mu _{k} mg d = \frac{1}{2} m v^{2}[/tex]
[tex]d = \frac{v^{2} }{2 g \mu _{k} }[/tex]
[tex]d = \frac{9}{2 \times 9.8 \times 0.250}[/tex]
[tex]d = 1.84[/tex] m
Therefore, the distance travel before stopping is 1.84 m