A 4.0-kg rock is dropped from a height of 12.0 m. At whatheight is the rock\'s k

Question

A 4.0-kg rock is dropped from a height of 12.0 m. At whatheight is the rock's kinetic energy twice its potentialenergy?_______m A sled and rider with a combined weight of 69kg are at rest onthe top of a hill 20m high. a) what is their total energy at the top of thehill_____J b)Assuming there is no friction, what would the total energybe on sliding halfway down the hill______J An object is dropped from a height of 13m. At what height willits kinetic energy and its potential energy be equal______m A 4.0-kg rock is dropped from a height of 12.0 m. At whatheight is the rock's kinetic energy twice its potentialenergy?_______m A sled and rider with a combined weight of 69kg are at rest onthe top of a hill 20m high. a) what is their total energy at the top of thehill_____J b)Assuming there is no friction, what would the total energybe on sliding halfway down the hill______J An object is dropped from a height of 13m. At what height willits kinetic energy and its potential energy be equal______m

Explanation / Answer

a) Just before drop the rock has only P.E = 4X10X12 (g =10m/s2)                                                                =480 J the rock falls freely under the gravitation, and given K.E. istwice P.E i.e. 1/2 mv2 = 2mgh'                h'= v2 /4g but from laws of body under gravitation,v2=2g(h-h') from above two equations h' = (h-h')/2 => h' = h/3 => h'=4m above the ground. As the sled and rider are at rest they possess only P.E. =69X10X20 (g = 10m/s2)                                                                                    =13800 J. In the half way they possess both P.E anD K.E P.E. = 69X10X10 = 6900 J. As the energy is conserved the K.E. = 6900 J given K.E. = P.E i.e. 1/2 mv2 = mgh'                h'= v2 /2g i.e. 1/2 mv2 = mgh'                h'= v2 /2g but from laws of body under gravitation,v2=2g(h-h') from above two equations h' = (h-h') => h' = h/2 => h'=6.5m above the ground. but from laws of body under gravitation,v2=2g(h-h') from above two equations h' = (h-h') => h' = h/2 => h'=6.5m above the ground.       

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