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Try this at home! PE and k
e) PEs in terms of k - Ideal Spring
1. P.E.s = Fx
Average Force: F = F/2
2. PE = (F/2)x
F = Kx
3. P.E.s = (1/2)(Kx)x
Ex 1) A force of 12. N stretches a spring and makes it .15m longer a) What's the spring constant (k) of this spring?
k = F/x
= 12. N/.15 m
= 80. N/m
b) P.E. of this spring?
PEs = (½)Kx2
= 1/2(80. N/m)(.15 m2)
PEs = .90 Joules
Ex 2) If 15. joules of energy are stored in the stretched spring, what is the value of the spring constant
15. J = (½)k(.50 m2)
15. J = .50 k(.25)
15 = .125 k
k = 120 N/m
W = DPE
29) A 1200 kg car rolling on a horizontal surface has a v = 65 km/hr when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring constant of the spring?
DKE = DPE 65 km/hr x [1000m/km][1hr/3600] = 18 m/s
1/2mv2 = 1/2kx2 mv2 = kx2 mv2/[x2] = k
8.1 x 104 N/m
©Tony Mangiacapre., -All Rights Reserved
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