Question:

A spring has a spring constant of 50 N/m. When a force of 10 N is applied to the spring, it stretches by a length of 0.2 m. Determine the potential energy stored in the spring when it is stretched by 0.4 m.

(A) 5 J (B) 10 J (C) 15 J (D) 20 J

Answer:

To solve this problem, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the distance the spring is stretched or compressed.

Hooke's Law is given by the formula:

F = -kx

Where: F is the force applied to the spring (in Newtons), k is the spring constant (in N/m), x is the displacement of the spring from its equilibrium position (in meters).

Given: k = 50 N/m F = 10 N (force applied to the spring) x = 0.2 m (displacement of the spring)

Let's calculate the potential energy stored in the spring when it is stretched by 0.2 m using the formula:

Potential energy (U) = 1/2 kx^2

Potential energy (U) = 1/2 * 50 N/m * (0.2 m)^2

Potential energy (U) = 1/2 * 50 N/m * 0.04 m^2

Potential energy (U) = 1 J

Now, let's determine the displacement of the spring when it is stretched by 0.4 m.

Since Hooke's law is a linear relationship, we can use proportions to find the new displacement.

Initial displacement (0.2 m) is to Initial force (10 N) as New displacement (0.4 m) is to New force (F).

0.2 m / 10 N = 0.4 m / F

Cross multiplying, we get:

0.2F = 4

F = 20 N

Now, let's calculate the potential energy stored in the spring using the new force and displacement:

Potential energy (U) = 1/2 * 50 N/m * (0.4 m)^2

Potential energy (U) = 1/2 * 50 N/m * 0.16 m^2

Potential energy (U) = 4 J

Therefore, the potential energy stored in the spring when it is stretched by 0.4 m is 4 J.

The correct answer is (D) 20 J.