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Dikerahui :
k = 1000 N/m
Δx = 10 cm = 0,1 m
Ditanyakan : m ?
Dijawab :
a. Paralel :
kp = k + 2k = 3 k = 3x1000 = 3000
massa :
F = kp Δx
mg = kp Δx
mx10 = 3000 x 0,1
mx10 = 300
m = 300/10
m =3 kg
b. seri :
1/ktot = 1/1000 + 1/(2x1000)
1/ktot = 1 / 1000 + 1/2000
1/ktot = (2+1) /2000
k tot = 2000/3
k tot = 666,67
jadi massa :
mg = ktot Δx
mx10 = 666,67 x 0,1
mx10 = 66,67
m = 66.67/10
m = 6,67 kg
Nomor 3 :
Diketahui :
k1 = k2 = 100 N/m
k3 = k4 = 200 N/m
Ditanyakan : a. K tot dan Δx jika F = 4.440 N
Dijawab :
hitung yg seri terlebih dahulu :
1/Ks = 1/k1 + 1/k2
1/Ks = 1/100 + 1/100
1/Ks = 2/100
Ks = 100/2
Ks = 50 N/m
Kparale :
Kp = 200 + 200 = 400 N/m
Ktotal :
1/ Ktot = 1/Ks + 1/Kp
= 1/50 + 1/400
= ((8 - 1) /400)
= 7/400
Ktot = 400/7
K tot = 57,14
Pertambahan panjang :
F = k Δx
Δx = F/k tot
= 4440/57,14
= 77.89
= 78 m