Question

If a spring stores 5 j of energy when its compressed by 0.5 m. what is the spring constant

Answer 1

Answer:

Spring constant = k = 10 N/m

Explanation:

According to Hooke's law:

                                         F = kx.

• F = The force applied to the spring in newtons (N).
• k = The spring constant measured in newtons per meter (N/m).
• x = The distance the spring is stretched from its equilibrium position in meters (m).

                                            k = F/x

Putting values in above formula:

                                            k = 5/0.5

                                             k = 10 N/m

Answer 2

The elastic potential energy stored in a spring is given by:

$E=\frac{1}{2}kx^2$

where k is the spring constant, and x=0.5 is the compression of the spring. Re-arranging the formula and using E=5 J, we find the spring constant:

$k=\frac{2E}{x^2}=\frac{2(5 J)}{(0.5 m)^2}=40 N/m$