gangelucci17
Answered

Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

If the spring constant is 12.6N/m and a spring is stretched by 0.25m, how much force has been applied?

Sagot :

Answer:

The magnitude of the applied force is 3.15 N

Explanation:

Hooke's law states that for small deformations, the change in dimensions, 'x', of an object is directly proportional to the force, 'F', of the acting load acting that bring about the deformation

Mathematically, Hooke's Law can be expressed as follows;

F = k·x

Where;

F = The applied load

k = The force constant of the object

x = The amount of deformation of the object

The given parameters are;

The spring's force constant, k = 12.6 N/m

The amount of stretching in the spring, x = 0.25 m

Therefore, we have;

The force applied, F = k·x = 12.6 N/m × 0.25 m = 3.15 N.

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.