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
