Respuesta :
Answer:
t = 1.62 s
Explanation:
given,
mass of the block m₁ = 16.5 Kg
m₂ = 8 Kg
angle of inclination = 60°
μs = 0.400 and μk = 0.300
time to slide 2 m = ?
a) let a is the acceleration of the block m₁ downward.
Net force acting on m₂,
F₂ = T - m₂ g
m₂a = T - m₂ g
[tex]a = \dfrac{T}{m_2} - g[/tex].......(1)
net force acting on m₁
F₁ = m₁g sin(60°) - μ_k m₁g cos (60°) - T
m₁ a = m₁g sin(60°) - μ_k m₁g cos (60°) - T
[tex]a = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}[/tex].........(2)
from equations 1 and 2
[tex]\dfrac{T}{m_2} - g = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}[/tex]
[tex]\dfrac{T}{m_2} +\dfrac{T}{m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)[/tex]
[tex]T\dfrac{m_1+m_2}{m_2\times m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)[/tex]
[tex]T = \dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{\dfrac{m_1+m_2}{m_2\times m_1}}[/tex]
[tex]T = {m_2\times m_1}\dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{{m_1+m_2}}[/tex]
[tex]T = {16.5\times 8}\dfrac{9.8 + 9.8 sin(60^0) - 0.3\times 9.8 cos (60^0)}{{16.5+8}}[/tex]
T = 90.61 N
from equation (1)
[tex]a = \dfrac{90.61}{8} - 9.8[/tex].......(1)
a = 1.52 m/s²
let t is the time taken
Apply,
d = ut + 0.5 a t²
2 = 0 + 0.5 x 1.52 x t²
[tex]t = \sqrt{2.63}[/tex]
t = 1.62 s
