Answer:
= 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,.
T - f_s - mg sin \theta = m aT−f
s
−mgsinθ=ma
T - \mu N - mg sin \theta = m aT−μN−mgsinθ=ma
T - \mu mg - mg sin \theta = m aT−μmg−mgsinθ=ma
a = \dfrac{T}{m} - \mu g - g sin \thetaa=
m
T
−μg−gsinθ
a = \dfrac{5000}{200} - \0.2\times 9.8 - 9.8 \times sin 30^0a=
200
5000
−\0.2×9.8−9.8×sin30
0
a = 18.14\ m/s^2a=18.14 m/s
2
L = \dfrac{10}{sin 30}L=
sin30
10
L = 20 m
v² = u² + 2 as
v² = 0 + 2 x 18.14 x 20
v = 26.94 m/s
