10 points A very light ideal spring with a spring constant (force constant) of 250
N/m pulls horizontally on an 18-kg box that is resting on a horizontal floor.
The coefficient of static friction between the box and the floor is 0.65,
and the coefficient of kinetic friction is 0.45. What length is the spring
stretched just as the box is ready to move? *
o
0.46 m
0.32 m
0.25 m
0.21 m

e = 0.46 m

Explanation:

From the laws of friction, frictional force, F is proportional to normal reaction, R.

F₁ = μR

where μ is coefficient of friction; R = mg and g = 9.8 ms⁻²

Also, from Hooke's law, extension, e, in an elastic spring is proportional to applied force.

F₂ = Ke

where K is force constant of the spring

Since the box is just about to move, the coefficient of friction involved is static friction.

The force on the spring equals the frictional force experienced by the box the box; F₁ = F₂

Ke = μR

e = μR/K

where μ = 0.65; R = 18 kg * 9.8 ms⁻²; K = 250 N/m

e = (0.65 * 18 * 9.8)/250

e = 0.46 m

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

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