Respuesta :
Answer:
v = 7.71 m/s
Explanation:
given,
angle of inclination = 22.9°
distance moved = 12.1 m
coefficient of friction = 0.15
final speed = ?
using work energy theorem
[tex]W = \dfrac{1}{2}mv^2[/tex]..........(1)
now,
v is the speed of the box.
W is the work done on the box by a net external force.
Work done on the force
[tex]W = F_{net}L[/tex]
[tex]F_{net} = m g sin \theta - \mu N[/tex]
[tex]F_{net} = m g sin \theta - \mu (mg cos \theta)[/tex]
[tex]W = (mg sin\theta - \mu mg cos \theta)L[/tex]......(2)
now, equating both equation 1 and 2
[tex]\dfrac{1}{2}mv^2 = (mg sin\theta - \mu mg cos \theta)L[/tex]
[tex]v= \sqrt{2 (g sin\theta - \mu g cos \theta)L}[/tex]
[tex]v= \sqrt{2(9.8\times sin 22.9^0 - 0.15\times 9.8 \times cos 22.9^0)12.1}[/tex]
v = 7.71 m/s
hence, the final speed of the box is equal to v = 7.71 m/s
