Respuesta :
Answer:
v = 26.94 m/s
Explanation:
given,
angle of inclination = 30°
mass of the sleigh = 200 kg
coefficient of kinetic friction = 0.2
height of inclination = 10 m
pull force be = 5000 N
now,.
[tex]T - f_s - mg sin \theta = m a[/tex]
[tex]T - \mu N - mg sin \theta = m a[/tex]
[tex]T - \mu mg - mg sin \theta = m a[/tex]
[tex]a = \dfrac{T}{m} - \mu g - g sin \theta[/tex]
[tex]a = \dfrac{5000}{200} - \0.2\times 9.8 - 9.8 \times sin 30^0[/tex]
[tex]a = 18.14\ m/s^2[/tex]
[tex]L = \dfrac{10}{sin 30}[/tex]
L = 20 m
v² = u² + 2 as
v² = 0 + 2 x 18.14 x 20
v = 26.94 m/s
