Question

A 16.5-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?

Answer 1

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

$a = \dfrac{T}{m_2} - g$.......(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

$a = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}$.........(2)

from equations 1 and 2

$\dfrac{T}{m_2} - g = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}$

$\dfrac{T}{m_2} +\dfrac{T}{m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)$

 $T\dfrac{m_1+m_2}{m_2\times m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)$

 $T = \dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{\dfrac{m_1+m_2}{m_2\times m_1}}$

 $T = {m_2\times m_1}\dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{{m_1+m_2}}$  

$T = {16.5\times 8}\dfrac{9.8 + 9.8 sin(60^0) - 0.3\times 9.8 cos (60^0)}{{16.5+8}}$

T = 90.61 N

from equation (1)

$a = \dfrac{90.61}{8} - 9.8$.......(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

$t = \sqrt{2.63}$

t = 1.62 s