A spring of spring constant 5*10 3 N/m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch is further by another 5 cm is ?

To find the work done in stretching the spring an additional 5 cm, we can use Hooke's Law and the formula for work done on a spring. Hooke's Law states that the force exerted by a spring is proportional to its displacement from the equilibrium position, which can be expressed as:

Understanding Hooke's Law

The formula is given by:

F = kx

where:

• F is the force exerted by the spring (in Newtons),
• k is the spring constant (in N/m), and
• x is the displacement from the equilibrium position (in meters).

Calculating Work Done on the Spring

The work done in stretching a spring can be calculated using the formula:

W = (1/2)kx²

In this scenario, the spring constant k is given as 5,000 N/m, and the spring is first stretched by 5 cm (0.05 m). Then, we stretch it further by another 5 cm, making the total stretch 10 cm (0.10 m). We need to calculate the work done during each of these stretches and then find the difference.

Step 1: Work Done for the Initial Stretch

For the initial stretch of 5 cm:

x₁ = 0.05 m

Using the work formula:

W₁ = (1/2) * k * x₁² = (1/2) * 5000 N/m * (0.05 m)²

W₁ = (1/2) * 5000 * 0.0025 = 6.25 J

Step 2: Work Done for the Total Stretch

Now, for the total stretch of 10 cm:

x₂ = 0.10 m

Calculating the work done for this total stretch:

W₂ = (1/2) * k * x₂² = (1/2) * 5000 N/m * (0.10 m)²

W₂ = (1/2) * 5000 * 0.01 = 25 J

Step 3: Finding the Work Done for the Additional Stretch

To find the work required to stretch the spring from 5 cm to 10 cm, we can subtract the work done for the first stretch from the work done for the total stretch:

Work for additional stretch = W₂ - W₁

Work for additional stretch = 25 J - 6.25 J = 18.75 J

Final Result

The work required to stretch the spring an additional 5 cm, after it has already been stretched by 5 cm, is 18.75 joules. This illustrates how the work done on a spring increases with the square of the displacement due to the nature of elastic potential energy. So, the more you stretch it, the more energy you need to apply! If you have any further questions about this concept or related topics, feel free to ask!