Question

Please help me

When a spring is compressed by a force F from point L to point K, as shown in the figure, its potential energy increases by 12 J. If the spring constant is 200 N/m find the distance, x. ​

Answer 1

The distance, x, given that the potential energy increased by 12 J, is 0.35 m

How to calculate the distance x?

The energy and distance (i.e compression or extension) of a spring are related according to the following equation:

`E = (1)/(2)Kx^2`

Where

• E is the energy stored in the spring in joule (J)
• K is the spring constant in newton per meter (N/m)
• x is the distance (i.e compression or extension) in meters (m)

From the question given, the spring constant and energy are given as:

• Spring constant (K) = 200 N/m
• Potential energy of spring (PE) = 12 J
• Distance (x) = ?

From the above information provided by the question, the distance, x can be calculated as follow:

`E = (1)/(2)Kx^2\\\\12 = (1)/(2)\ *\ 200\ *\ x^2\\\\12 = 100\ *\ x^2\\\\Divide\ both\ sides\ by\ 100\\\\x^2 = (12)/(100) \\\\Take\ the\ square\ root\ of\ both\ sides\\\\x = \sqrt{(12)/(100)} \\\\x = 0.35\ m`

Thus, we can conclude that the distance, x is 0.35 m