A force of magnitude 800N stretches a certain spring by 0.200mform its equilibrium position. a) what is the force constant of thisspring? k= N/m b) how much elastic potential energy is stored in the springwhen it is stretched 0.300m from its equilibriumposition.? U = J c) how much elastic potential energy is stored in the springwhen it is compressed by 0.300m from its equilibrium? U = J d) how much work was done is streching the spring by theoriginal 0.200m? W = J A force of magnitude 800N stretches a certain spring by 0.200mform its equilibrium position. a) what is the force constant of thisspring? k= N/m b) how much elastic potential energy is stored in the springwhen it is stretched 0.300m from its equilibriumposition.? U = J c) how much elastic potential energy is stored in the springwhen it is compressed by 0.300m from its equilibrium? U = J d) how much work was done is streching the spring by theoriginal 0.200m? W = J
Explanation / Answer
(a): F = kx 800 = k(0.200) k = 800/0.200 k = 4000 N/m -- (b) U = 1/2kx^2 U = (0.5)(4000)(0.300)^2 U = 180 J -- (c) same distance in different direction, so same magnitde indifferent direction = 180 J -- (d) Work = F*d Work = (0+800)/2*0.200 Work = 80 J -- Hope this helps.