Question

A brick is given an initial speed of 5 ft/s up an inclined plane at an angle of 30° from the horizontal. The coefficient of friction is u = V3/12. After 0.5 seconds, how far is the brick from its original position? You may take g= 32 ft/s”, sin 30º = 1/2, and cos 30º = V3/2.

Answer 1

N 14 AN I mg sino nalost ing the masse -L. after 0.5 Seconds, distance covered by net force on mass Emet = lng Sint tfs) mig sino tu glost). fs= M N = ma Ming losse N= mglossa M -32 x į t 42 a=-g(Sino +of lost) aceleration ( t. 3 x 2*2) a= -32 (1+3 - 1685 4 a = -2oft/s? Selve u² thas) (s=ut + taty 2

u= utat let final velocity be zero att seconds. 0 =5-20t (t= 0.25 See) distance covered in 0.25 Sec. w²u²+zas u? - 25 -(0.625ft) s za 2 x 20 Nous, mass will start sliding back Angel Na f Sint Forma loot ma at ing az g (Sint-M-lost). a z 32x c 4 12 mis n's(-3) =Hx3 f

distance covered in 0:25 See saut + + 1 gt? 4x12x(0.25)? a (S = 0.375 0ft) to 628 ft 0.375ft 2-AS 0.625 -0.375 AS =5,-& (45= 0.25 ft)