Question

Q3 Figure 3.1 shows a 100 kg crate starts from rest and achieves a velocity of v = 3 m/s when it travels a distance of 6 m to the right. Determine the magnitude of force P acting on the crate if: (a) the coefficient of kinetic friction between the crate and the floor is 0.35. (15 marks) (b) the contact between the crate and the floor is smooth. (10 marks) P 30° Figure 3.1: A crate moved by force P

Answer 1

Solution Problem a opsin 30 N P FBD 30 e Pcos 30 UN mg For equilibrium : N+ Psin 30 = ng ( vertical eqm) - Ns m4 - Phin 30 and we know; F =ma Pcos 30 - UN = ma ( Horizontal equilibrium P cos30 - UN a= 2 m

as we know va =u²+zas (uzo) V2 v? - - २ = 2 [Passo - um seinen = { posa y/(wy-come). -(2) -finale 2 PCOS30 - 0.35 (100X9.31 – Psinso) 3 = 2 X6 100 9 X 100 PCOS 30 - 0.35(981. - PSin 30) 2 x 6 75 PCOs30 - 343.35 +0.175P - 418.35 = P(Cos 30 + 0.175) P = 401.86 N Ans

Problem b ! As surface is smooth (l=o) PCOS 30 - UN = ma .. ma= Pcos 30 (u=) d = pCosso 3 Now; V? t zas as P cos30 V2= २ [ S 32 = 2 P cos30 100 -7x X 6 9x100 2 X6 p cos30 a vo P = 86.602 N Ans J