A force of 10N extends a spring of natural length 1m by 0.02m, calculate the length of the spring when the applied force is 40N.

Answer Details

To solve this problem, we will use Hooke's Law. Hooke's Law states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it is represented as:

F = k * x

where:

• F is the force applied to the spring.
• k is the spring constant.
• x is the extension or compression of the spring from its natural length.

Firstly, we need to find the spring constant k. We know that a force of 10N extends the spring by 0.02m. Therefore, using Hooke's Law:

10N = k * 0.02m

From this, we can solve for k:

k = 10N / 0.02m = 500N/m

Now that we have determined the spring constant, let's calculate the extension caused by a force of 40N:

Using Hooke's Law again:

F = k * x

40N = 500N/m * x

Solving for x:

x = 40N / 500N/m = 0.08m

This means that the spring is extended by 0.08m when a force of 40N is applied. Therefore, the length of the spring (natural length plus extension) becomes:

1.00m + 0.08m = 1.08m

Thus, the **length** of the spring when the applied force is 40N is 1.08m.