A steel spring is hung vertically. Using a ruler, the lower end of the spring was measured to be 25.0cm from its upper end. However, when an object of mass 0.500kg it attached to the lower end of the spring the spring stretches to a new length of 45.0cm.

With the aid of a diagram and with a detailed explanation of your reasoning determine:

(a) The stiffness constant k of the spring.

(b) The additional weight needed to extend the spring to 50.0cm.

(c) The force needed to extend the spring by 50.0mm from its natural length.

See previous post: Thu, 11-17-16, 6:05 AM.

To determine the stiffness constant of the spring (k), we can use Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the displacement from its equilibrium position.

(a) The formula used to calculate the stiffness constant is:

F = k * Δx

Where:
F is the force applied to the spring
k is the stiffness constant
Δx is the displacement from the equilibrium position

In this case, the spring is initially at 25.0cm (0.25m) and stretches to 45.0cm (0.45m) when the object of mass 0.500kg is attached.

To find the force (F), we need to consider the weight of the object, which is given by:

Weight = mass * gravitational acceleration

Taking the gravitational acceleration to be approximately 9.8 m/s^2, we have:

Weight = 0.500kg * 9.8 m/s^2 = 4.9 N

The force exerted on the spring is equal to the weight of the object, so F = 4.9 N.

Now, we can calculate the displacement (Δx) from the equilibrium position:

Δx = 0.45m - 0.25m = 0.20m

Now we can substitute the values into the formula and solve for k:

4.9 N = k * 0.20m
k = 4.9 N / 0.20m = 24.5 N/m

Therefore, the stiffness constant (k) of the spring is 24.5 N/m.

(b) To determine the additional weight needed to extend the spring to 50.0cm (0.50m), we can use the same formula:

F = k * Δx

Considering the new displacement (Δx) from the equilibrium position:

Δx = 0.50m - 0.25m = 0.25m

We already know that k = 24.5 N/m. Now we can calculate the force (F) required:

F = 24.5 N/m * 0.25m = 6.125 N

Since the original weight was 4.9 N, we need an additional force of:

Additional force = 6.125 N - 4.9 N = 1.225 N

Therefore, the additional weight needed to extend the spring to 50.0cm is 1.225 N.

(c) To determine the force needed to extend the spring by 50.0mm (0.050m) from its natural length, we can once again use Hooke's Law:

F = k * Δx

Considering the new displacement (Δx) from the natural length:

Δx = 0.050m

We already know that k = 24.5 N/m. Now we can calculate the force (F) needed:

F = 24.5 N/m * 0.050m = 1.225 N

Therefore, the force needed to extend the spring by 50.0mm from its natural length is 1.225 N.