Respuesta :
Answer:
The distance travel by block before coming to rest is 0.122 m
Explanation:
Given:
Mass of block [tex]m = 6.60[/tex] kg
Initial speed of block [tex]v _{i} = 1.56[/tex] [tex]\frac{m}{s}[/tex]
Final speed of block [tex]v_{f} = 0[/tex] [tex]\frac{m}{s}[/tex]
Coefficient of kinetic friction [tex]\mu _{k} = 0.62[/tex]
Ramp inclined at angle [tex]\theta =[/tex] 28.4°
Using conservation of energy,
Work done by frictional force is equal to change in energy,
[tex]\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U[/tex]
Where [tex]\Delta U = mg d\sin 28.4[/tex]
[tex]\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4[/tex]
[tex]\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}[/tex]
[tex]d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}[/tex]
[tex]d = 0.122[/tex] m
Therefore, the distance travel by block before coming to rest is 0.122 m
