It takes 10 lb of force to stretch a spring 2 in. What is the stiffness (spring constant) of the spring

875.59 N/m

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

c) up is the direction the magnetic force is acting on the charge.