Answer:
875.59 N/m
Step-by-step explanation:
From the question given above, the following data were obtained:
Force (F) = 10 lb
Extention (e) = 2 in
Spring constant (K) =?
Next, we shall convert 10 lb to Newton (N). This can be obtained as follow:
1 lb = 4.448 N
Therefore,
10 lb = 10 lb × 4.448 N / 1 lb
10 lb = 44.48 N
Thus, 10 lb is equivalent to 44.48 N
Next, we shall convert 2 in to metre (m). This can be obtained as follow:
1 in = 0.0254 m
Therefore,
2 in = 2 in × 0.0254 m / 1 in
2 in = 0.0508 m
Thus, 2 in is equivalent to 0.0508 m.
Finally, we shall determine the spring constant i.e the stiffness of the spring as follow:
Force (F) = 44.48 N
Extention (e) = 0.0508 m.
Spring constant (K) =?
F = Ke
44.48 = K × 0.0508
Divide both side by 0.0508
K = 44.48 / 0.0508
K = 875.59 N/m
Therefore the spring constant i.e the stiffness of the spring is 875.59 N/m