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Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a

force constant of 150 N/m. There is simple friction between the object and surface with a static
coefficient of frictions 0.100μ = . (a) How far can the spring be stretched without moving the
mass? (b) If the object is set into oscillation with an amplitude twice the distance found in part
(a), and the kinetic coefficient of friction isk 0.0850μ = , what total distance does it travel before
stopping? Assume it starts at the maximum amplitude

1 Answer

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Final answer:

The spring can be stretched approximately 0.049 meters without moving the mass. If the object is set into oscillation with an amplitude twice this distance, it will travel a total distance of 0.098 meters before stopping.

Step-by-step explanation:

To determine how far the spring can be stretched without moving the mass, we need to consider the maximum force that the static friction can exert on the object. The maximum static friction force is given by

fs = μs * mg,

where μs is the static coefficient of friction and mg is the weight of the object.

The force exerted by the spring is given by

Fspring = k * x,

where k is the force constant of the spring and x is the distance the spring is stretched.

When the object is on the verge of moving, the static friction force is equal to the force exerted by the spring, so we have fs = Fspring.

Plugging in the values, we have:

μs * mg = k * x

Solving for x, we get x = (μs * mg) / k. Substituting the given values, we have:

x = (0.100 * 0.750 * 9.8) / 150

x ≈ 0.049 m

Therefore, the spring can be stretched by approximately 0.049 meters without moving the mass.

In part (b), when the object is set into oscillation with amplitude twice the distance found in part (a), the total distance it will travel before stopping can be calculated using the formula d = 2 * A, where A is the amplitude.

Therefore, the total distance travelled is

d = 2 * 0.049 = 0.098 m.

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