An elastic cord can be stretched to it elastic limit by a load of 2N if a 35cm length of cord is extend to 0.6cm by a force of 0.5N . What will be the length of the cord when it is stretching cord is 2.5N

Answer 1

To determine the relationship between the force applied to the cord and the amount of stretch, we can use Hooke's Law. Hooke's Law states that the force applied to a spring (or elastic cord) is directly proportional to the amount of stretch or compression.

In this case, we are given that a load of 2N stretches a 35cm length of cord to 0.6cm. We can use this information to find the proportionality constant, or the spring constant (k), using the formula:

k = (F / x)

where F is the force applied and x is the amount of stretch.

k = (2N / 0.6cm) = 3.33 N/cm

Now, we can use this spring constant to find the length of the cord when a force of 2.5N is applied. Let L1 be the original length of the cord and L2 be the stretched length of the cord when a force of 2.5N is applied.

k = (F / x)
3.33 N/cm = (2.5N / (L2 - L1))

We know the original length (L1) is 35cm and the force applied (F) is 2.5N. Plugging in these values, we can solve for L2:

3.33 N/cm = (2.5N / (L2 - 35cm))

Cross-multiplying, we get:

3.33 N/cm * (L2 - 35cm) = 2.5N

Dividing both sides by 3.33 N/cm:

(L2 - 35cm) = 2.5N / 3.33 N/cm

Simplifying, we have:

L2 - 35cm = 0.7515 cm

Adding 35cm to both sides:

L2 = 35.7515 cm

Therefore, the length of the cord when it is stretched by a force of 2.5N is approximately 35.7515 cm.