A16.5-kg crate starts at rest at the top of a 60.0° incline. the coefficients of friction are μs = 0.400 and μk = 0.300. the crate is connected to a hanging 8.00-kg box by an ideal rope and pulley. how long does it take the crate to slide 2.00 m down the incline?

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

.......(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

.........(2)

from equations 1 and 2

T = 90.61 N

from equation (1)

.......(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

t = 1.62 s

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it is a negative acceleration