Force F on the particle of mass 0.1 kg varies with distance x as shown in figure . If particle with distance x as shown in figure .If particle move from rest x = 0 , that at x = 12 cm , what will be its velocity ?

Explanation

Work W = Area of isoscles trapezoid OABC
` =1/2 {:{("sum of sides"),(of " parallelogram"):}}xx{("perpendicular distance"),("between sides of parallelogram"):}}`
` =1/2 (AB +OC) xx(BD)`
` =1/2 (4+12)xx(10)`
` = 80 J `
`rArr` Now according to work energy theorem ,
`W =K_(f) -K_(0)`
` :. 80 =1/2 mv_(f)^(2) -1/2mv_(i)^(2)`
` :. 80 =1/2 mv_(f)^(2) " " [ :. v_(i) = 0] `
` 80 =1/2 xx0.1 xxv^(2)`
` :. 1600 =v^(2)`
` :. v = 40 m^(-1)`