Respuesta :
Answer:
A) 35.5N/m b) 20.1cm
Explanation:
Using Hooke's law;
F = Ke where F is the weight of the object = mass of the object in kg * acceleration due to gravity in m/s^2 and k if the force constant of the spring in N/m and e is the extension of the spring which original length of the spring - new length after extension in meters
For the first body, m*g = K * (0.25- li)
Where li is the initial length of the spring
0.175*9.81 = k(0.25-li)
1.72 = k(0.25-li) as equation 1
For the second body, m *g = K* ( 0.775-li)
2.075*9.81 = k (0.775-li) equation 2
20.36 = k(0.775-li)
Make li subject of the formula;
li = 0.775 - 20.36/k
Substitute for li in equation 1
1.72 = k(0.25- (0.775 - 20.36/k))
1.72 = k ( 0.25 - 0.775 + 20.36/k)
Open the bracket with k
1.72 = 0.25k - 0.775k + 20.36 (since k cancel k)
Collect the like terms:
1.72 - 20.36 = - 0.525k
- 18.64 = -0.525k
Divide both side by -0.525
-18.64/-0.525 = -0.525/-0.525k
K = 35.5N/m
B) substitute for k in using
li = 0.775 - 20.36/k
li = 0.775 - 20.36/35.5
li = 0.775 - 0.574
li = 0.201 in meters
li = 0.201 * 100 centimeters = 20.1cm
(A) The force constant of the spring is obtained as 34.66 N/m
(B) The unloaded length of the spring is 20 cm.
Hooke's Law
(A) According to Hooke's law;
[tex]F = kx[/tex]
Where 'F' is the force needed to extend or compress the spring, 'k' is the force constant and 'x' is the extended length or compressed length.
Here, the gravitational force of the mass provides the necessary force for the spring to elongate.
Therefore, we can say that in the case of first mass;
[tex]k\,(0.25-l_0) = m_1g = (0.175\,kg \times 9.8\,m/s^2)[/tex]
Where [tex]l_0[/tex] is the unloaded length of the spring.
[tex]\implies k\,(0.25-l_0)=1.715\,N[/tex]
In the case of the second mass, we can write;
[tex]k\,(0.775\,m-l_0) = m_2g = (2.075\,kg \times 9.8\,m/s^2)[/tex]
[tex]\implies k\,(0.775\,m-l_0) = 20.335\,N[/tex]
From both these equations, we can write;
[tex]k\,(0.775\,m-l_0-0.25\,m+l_0)=20.335\,N- 1.715\,N[/tex]
[tex]\implies0.525\, k=18.2\,N[/tex]
[tex]\therefore k=\frac{18.2\,N}{0.525\,m}=34.66\,N/m[/tex]
(B) Applying the value of 'k' in any of the equations for force of given masses, we get;
[tex]k\,(0.25-l_0)=1.715\,N[/tex]
[tex]\implies (34.66\,N/m)\times\,(0.25-l_0)=1.715\,N[/tex]
[tex]\implies 8.665N-34.66\,l_0=1.715\,N[/tex]
[tex]\implies \,l_0=\frac{-6.95N}{-34.66} =0.2\,m=20\,cm[/tex]
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