# A 400N bag of cement starts from rest and slides 12m down a roof with slope of 3 horizontal to 1 vertical and a coefficient ..?

Cont..

of friction of 0.20. If the vertical distance from the edge of roof to the ground is 10 m, how far from the wall will the bag of cement strike the ground?

### 2 Answers

- FiremanLv 72 months agoFavourite answer
Given: W = mg = 400 N = 400/9.8 kg = 40.82 kg & tanθ = 1/3

=>sinθ = 1/√{(3)^2 + (1)^2} = 1/√10 = 0.32

=>cosθ = 3/√{(3)^2 + (1)^2} = 3/√10 = 0.95By F(net) = F(gravity) - F(friction)=>ma = mgsinθ - µk x N=>ma = mgsinθ - µk x mgcosθ=>ma = W(sinθ - µk x cosθ)=>40.82 x a = 400(0.32 - 0.20 x 0.95)

=>a = 1.28 m/s^2

Thus by v^2 = u^2 + 2as

=>v^2 = 0 + 2 x 1.28 x 12

=>v = √30.64

=>v = 5.53 m/s

Now v(X) = vcosθ = 5.53 x 0.95 = 5.26 m/s

& v(Y) = vsinθ = 5.53 x 0.32 = 1.77 m/s

Let the bag take t sec to cover 10 m under v(Y), by s = ut + 1/2gt^2

=>10 = 1.77 x t + 1/2 x 9.8 x t^2

=>4.9t^2 + 1.77t - 10 = 0

=>t = [-1.77 +/-√{(1.77)^2 - 4 x 4.9 x (-10)}]/(2 x 4.9)

=>t = 1.26 sec

Thus distance covered on X-axis (Let R) under v(X), by s = ut

=>R = v(X) x t

=>R = 5.26 x 1.26 = 6.62 m ----------The required Answer.

- oubaasLv 72 months ago
mass =400/9.80665 = 40.79 kg

tan Θ = 1/3 ;

Θ = 18.435 ° ; sin Θ = 0.316 ; cos Θ = 0.949

acceleration a = 400*(0.316- 0.949*0.20)/40.79 = 1.238 m/sec^2

V = √ 2ad = √ 2*12*1.238 = 5.45 m/sec

Vy = -5.45*0.316 = -1.72 m/sec

-12 = -1,72*t-9,80665*t^2

falling time t = (1,72-√ 1.72^2+48*9.80665)/-19,613 = 1.022 sec

distance d = 5.45*0.949*1.022 = 5.28 m

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But why 10 m sir? The problem says its 12 m