Question

A force of 550 N keeps a certain ideal spring stretched at a distance of 0.600 m.

A) What is the potential energy of the spring when it is stretched 0.600 m?
B) What is its potential energy when it is compressed 8.00 cm?

Answer 1

The potential energy stored in a spring when it is compressed 8.00 cm, given an initial force of 550 N at 0.600 m stretch, is calculated to be 2.933 joules using Hooke's Law and the formula for potential energy in a spring (U = 1/2 k x²).

Step-by-step explanation:

The question asks about the potential energy stored in a spring when it is compressed. If a force of 550 N keeps a spring stretched at 0.600 m, we can use Hooke's Law to calculate the spring constant 'k', which is given by the formula k = F/x, where 'F' is the force applied and 'x' is the distance the spring is stretched or compressed from its natural length. Then, to find the potential energy 'U' when the spring is compressed 8.00 cm (which is 0.080 m), we use the formula for potential energy in a spring: U = 1/2 k x².

First, calculate the spring constant 'k':

k = 550 N / 0.600 m = 916.67 N/m

Now, determining the potential energy when the spring is compressed by 0.080 m:

U = 1/2 * 916.67 N/m * (0.080 m)²

U = 1/2 * 916.67 N/m * 0.0064 m²

U = 2.933 J

Therefore, the potential energy stored in the spring when it is compressed 8.00 cm is 2.933 joules.