Question

We assumed that the spring is ideal. In reality, as you stretch and compress the spring, frictional forces of the spring come into play. Explain how the frictional forces of the spring would or could affect your measurements.

Answer 1

Hi buddy!

Let me explain to you how the frictional forces of a spring affect our measurements while compressing or stretching!

By using Hooke's law, we know that the force exerted while stretching or compressing a spring is given by

F = - k x

The potential energy stored by a compressed spring is given by U_s =

gravitational potential energy of an object at a height h, above an arbitrarily defined zero point is

Ug=mgh

m = mass of object and g = acceleration due to gravity 9.8 m/s^2

Forces due to friction causes a significant effect in this case, so we will try to account for it. The frictional force on an object moving horizontally is

F_fx = μmg =ma

µ is the coefficient of friction and a is the acceleration. Since acceleration is the time derivative of velocity, and we can cancel the mass above. Thus the above equation can be written as,

while on an incline plane, the frictional force is given by

F_f =μmgcosθ

= angle of slope/incline

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