+244 A spring is initially stretched 0.75 m from equilibrium. It takes you 100 J of energy to stretch it further away from equilibrium to 1.25 m. What is the spring constant (force constant) of this spring?
500 N/m
800 N/m
50 N/m
200 N/m
+37146 Spring constant F = kx and we know (or should know) F = ma (Newtons) and ONE joule = ONE Newton-Meter
Stretch distance = 1.25 - .75 = .5 m
100 j = 100 N-m
F= 100 N-m / .5 m = 200 N
F= kx
200 = k(.5)
200N / .5m = k = 400 N/m ?????? Hmm...... second time today that I got an answer that does not match your variable choice answers..........I do not know what to think..... At what answer did YOU arrive???
ANYONE???
+33661 The energy required to stretch the spring away from equilibrium by distance x is given by (1/2)kx^2, so, in this case:
(1/2)k(1.25^2 - 0.75^2) = 100
k(25/16 - 9/16) = 200
k = 200 N/m. (Since 25/16 - 9/16 = 1)
(Note that the energy is the integral of force with respect to distance here, as it becomes increasingly difficult to stretch the spring the further it is away from equilibrium).
