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Title: Potential and Kinetic Energy Problem-high school physics? Post by: lgorman20101 on Jun 5, 2012 A constant force of 15.3 N acts upward on a iron block that weighs 11.7 N. (a) If the block is initially at rest, what is its kinetic energy 6.53 s after the force is applied? What is the increase in the potential energy of the body 6.53 s after the force is applied.
Title: Potential and Kinetic Energy Problem-high school physics? Post by: o.avworo on Jun 5, 2012 Given:
Fa = 15.3 N upward Fg = 11.7 N downward Find: a) K.E. = ? at t = 6.53 sec and Vi = 0 b) delta P.E. = increase in potential energy Solution: a) K.E. = (1/2)mv^2 K.E. = (1/2)(Fg/g)v^2 K.E. = (1/2)(11.7/9.8)v^2 Solving for v, final velocity: Fnet = ma Fa - Fg = (Fg/g)(a) a = ((Fa - Fg)/(Fg/g) a = (15.3 - 11.7)/(11.7/9.8) a = 4.3 m/s^2 a = (v - vi)/t v = vi + at v = 0 + (4.3m/s^2)6.53s v = 28.1 m/s Hence, K.E. = (1/2)(11.7/9.8)v^2 K.E. = (1/2)(11.7/9.8)(28.1)^2 K.E. = 471 joules ANS. b) delta P.E.g = mgh delta P.E.g = (11.7)(h) Solving for h: 2ah = v^2 - vi^2 h = (v^2 - vi^2)/2a h = (28.1^2 - 0)/2(4.3) h = 91.8 m delta P.E.g = (11.7)(91.8) delta P.E.g = 1074.8 joules teddyboy |