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
         = 2,5×105/1,6×10-2
 = 1,56×107 N/m2
5.      Given              : k= 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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