Soal dan Pembahasan Elastisitas (Fisika Kelas XI)
Elasticity
1.
A spring when pulled by a force of 20 N
increased length of 6 m. What is the potential energy?
A. 1,2
x 102 N/m
B. 2,33
x 102 N/m
C. 3,33 × 102 N/m
D. 4,2
x 102 N/m
E. 5,22
x 102 N/m
2.
A bar of steel which is 4 mm2
in cross-sectional area and 4 cm in length is pulled by a force of 100 N. If
the modulus of elasticity of steel is 2 x 1011 N/m2, so
the stress and strain of the steel is. . .
A. 1,25
x 104 N/m2 and 2,5 x 107
B. 1,25
x 10-4 N/m2 and 2,5 x 105
C. 2,5
x 105 N/m2 and 1,25 x 104
D. 2,5
x 106 N/m2 and 1,25 x 10-4
E. 2,5
x 107 N/m2 and 1,25 x 10-4
3.
A cylinder made of steel has a length of
10 m and diameter 4 cm. What is the length increment of the cylinder if it is
given a load of 105 N. (E = 2 x 1011 N/m2)?
A. 3,89
x 10-4 m
B. 3,98
x 10-3 m
C. 3,89
x 103 m
D. 3,88
x 104 m
E. 3,98
x 103 m
4.
A strand of wire has a length of 50 cm
with a cross-sectional area 2 cm2. A force of 50 N work on the wire
so the wire length increases 0,8 cm. The modulus elasticity is. . .
A. 1,56×107 N/m2
B. 1,57 x 108 N/m2
C. 1,66 x 107 N/m2
D. 1,67 x 107 N/m2
E. 1,77 x 108 N/m2
5.
Three springs in series, each spring has
constants of 1200 N/m, 600 N/m, 400 N/m. Three springs is given force by 40 N.
What is the spring constants replacement?
A. 100
N/m D. 800 N/m
B. 200
N/m E. 1200 N/m
C. 400
N/m
KEY
ANSWER :
1.
C. 3,33 × 102 N/m
2.
E. 2,5 x 107 N/m2
and 1,25 x 10-4
3.
B. 3,98 x 10-3 m
4.
A. 1,56×107 N/m2
5. B.
200 N/m
SOLUTION
:
1.
Given :
F = 20 N
l
= 6 cm
Ask :
Ep. . .
Answer : F = 20 Nx = 6 cm = 6×10-2 m
k = F/x
k = 20/6×10-2
k = 3,33×102 N/m
2.
Given : A = 4 mm2 = 4 x 10-6
L
= 40 cm
F
= 100 N
E
= 2
x 1011 N/m2
Ask :
Stress and Strain. . .
Answer : Stress = F/A
= 100/(4 x 10-6) = 2,5 x 107
N/m2
Strain = Stress/E
= (2,5 x 107)/
(2 x 1011) = 1,25 x 10-4
3.
Given : l = 10 m
D
= 4 cm
F
= 105 N
E
= 2
x 1011 N/m2
A=
¼ π d2 = 1,256 x 10-3 m2
Ask : ∆l. . .
Answer : ∆l =
(F x L)/(E.A)
=
(105 N x 10 m)/ ((2 x 1011 N/m2)
x (1,256 x 10-3 m2)
=
3,98 x 10-3 m.
4.
Given : l1 = 50 cm
A
= 2 cm2
F
= 50 N
∆l
= 0,8
Ask : Modulus Elasticity. . .
Answer :
E = Stress/strain
E =
2,5×105/1,6×10-2
= 1,56×107 N/m2
5.
Given : k1 = 1200 N/m
k2 = 600 N/m
k3 = 400 N/m
Ask : constants
replacement. . .
Answer :
1/kp = 1/k1 + 1/k2 + 1/k3
=
1/1200 + 1/600 + 1/400
=
6/1200
Kp = 200 N/m
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