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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.

User Mcoav
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1 Answer

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Final answer:

The maximum distance the spring can be stretched without moving the mass is approximately 0.0491 m. The total distance the object travels before stopping is approximately 0.3928 m.

Step-by-step explanation:

(a) How far can the spring be stretched without moving the mass?

To determine the maximum distance the spring can be stretched without moving the mass, we need to consider the balance between the force from the spring and the static friction force. The maximum static friction force is given by:

fs = μsmg

where μs is the static coefficient of friction, m is the mass, and g is the acceleration due to gravity. In this case,

μs = 0.100, m = 0.750 kg, and g = 9.8 m/s2.

Therefore, the maximum distance the spring can be stretched without moving the mass is:

x = fs/k

where k is the force constant of the spring. In this case, k = 150 N/m. Substituting the values:

x = (0.100)(0.750 kg)(9.8 m/s2)/(150 N/m)

x ≈ 0.0491 m

(b) What total distance does it travel before stopping?

To calculate the total distance the object travels before stopping, we need to consider the balance between the force from the spring, the kinetic friction force, and the damping force due to air resistance. The damping force can be ignored in this case, as it is not mentioned in the question.

The total distance traveled can be calculated as the sum of the distances traveled during each oscillation. Each oscillation consists of two equal halves: the distance from the maximum amplitude to the equilibrium position (twice the distance found in part (a)), and the distance from the equilibrium position to the maximum amplitude.

Therefore, the total distance traveled is:

d = 2x + 2y

where y is the distance from the equilibrium position to the maximum amplitude, which is equal to the distance found in part (a).

Substituting the values:

d = 2(2)(0.0491 m) + 2(0.0491 m)

d ≈ 0.3928 m

User Ratojakuf
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