Pls explain in DETAIL the FORMULA and EQUATIONS and how did you come up with the solution. I need to present it in class and I didn't know how 6.Given:Mass of block =MAngle of incline = 0Length of incline =SHeight of incline hHeight(h) = S sin eS=Force equation in direction vertical to Incline:N - Mg cos e = M x0[Since there is no motion vertically, so accleration = 0]N = Mg cos e .... EquationForce equation in direction horizontal to Incline:uN – Mg sin e = M x -a[Since there is motion horizontally down, so accleration = -a]U sin g equation (1).u (Mg cos 0) - Mg sin e = M X(-a)- a- PIMg cos -Me sinMMgia cos 8-sin e)- a =a = g(sin - H cos 0)a = g sin e (1 - u s)sina = g sin o(1 - pcot 0)... Equation(2)U sin g 3rd equation of motion12 = u? + 2aswhere,v = final velocityu = initial velocitya = accelerationS = displacementU sin g equation (2).2 = 0 + 2(g sin e(1 - pcot 0)Sv = V2g sin 0(1 - pcot 0)SAlso, sin ce S =sinv = V2g sin 0(1 – ucot 0)sin ev = V2gh(1- pcot 6)Change in Gravitational Potential Energy:Gravitional Potential Energy = mass x height x acceleration due to gravityU = mghChange in Potential energy = Potential energy at height h - Potential energyat height 0AU = Mgh – Mg x0AU = MghorMgSAU =sin 8[Since S = h sin e] Work Done by frictional Force:F = uNwhere,F = Force of frictionH = coefficient of frictionN = normal forceU sin g equation (1).F = uMg cos eWork Done = / Fdxwhere,F = forcex = displacementw = / µMg cos 0 x (-dx)IS p is varying with x,W = -Mg cos 0 / µ(x)dx

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Given: Mass of block =M Angle of incline 0 Length of incline S Height of incline = h Height(h) - S sin o S = Force equation in direction vertical to Incline : N- Mg cos ở = M X0 [Since there is no motion vertically, so accleration = 0] N = Mg cos e .. Equation(1) Force equation in direction horizontal to Incline : uN - Mg sin o = M x -a [Since there is motion horizontally down, so accleration = -a] U sin g equation (1). H(Mg cos ) - Mg sin 0 = MX(-a) - a EIME CIn -My sin M Mela cos -sin d a = g (sin - pı cos 0) a- g sin o (1- S sin a- g sin 0(1 - pcot 0) ..... Equation(2) U sin g 3rd equation of motion v2 = u? + 2as where. V final velocity - initial velocity a = acceleration S = displacement U sin g equation (2). =0 + 2 (g sin e(1 - pcot 0)S v = 2g sin 0(1 - µcot 0)S Also, sin ce S -. vV2g sin 0(1 - pcot 0) v- /2gh(1- pcot 0) Change in Gravitational Potential Energy: Gravitional Potential Energy = mass x height x acceleration due to gravity U = mgh Change in Potential energy = Potential energy at height h - Potential energy at height 0 AU = Mgh - Mg x 0 AU = Mgh or MgS AU = [Since S = h sin o sin 6

Given: Mass of block =M Angle of incline 0 Length of incline S Height of incline = h Height(h) - S sin o S = Force equation in direction vertical to Incline : N- Mg cos ở = M X0 [Since there is no motion vertically, so accleration = 0] N = Mg cos e .. Equation(1) Force equation in direction horizontal to Incline : uN - Mg sin o = M x -a [Since there is motion horizontally down, so accleration = -a] U sin g equation (1). H(Mg cos ) - Mg sin 0 = MX(-a) - a EIME CIn -My sin M Mela cos -sin d a = g (sin - pı cos 0) a- g sin o (1- S sin a- g sin 0(1 - pcot 0) ..... Equation(2) U sin g 3rd equation of motion v2 = u? + 2as where. V final velocity - initial velocity a = acceleration S = displacement U sin g equation (2). =0 + 2 (g sin e(1 - pcot 0)S v = 2g sin 0(1 - µcot 0)S Also, sin ce S -. vV2g sin 0(1 - pcot 0) v- /2gh(1- pcot 0) Change in Gravitational Potential Energy: Gravitional Potential Energy = mass x height x acceleration due to gravity U = mgh Change in Potential energy = Potential energy at height h - Potential energy at height 0 AU = Mgh - Mg x 0 AU = Mgh or MgS AU = [Since S = h sin o sin 6

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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