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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 friction μs=0.100 .

(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 is μk=0.0850 , what total distance does it travel before stopping? Assume it starts at the maximum amplitude.

2 Answers

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Final answer:

To find the maximum distance the spring can be stretched without moving the mass, calculate the force of static friction and equate it to the spring force.

Step-by-step explanation:

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

To find the maximum distance the spring can be stretched without moving the mass, we need to calculate the force of static friction. The force of static friction can be calculated using the equation:

Fs = μs * m * g

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

Once we have the force of static friction, we can equate it to the spring force:

Fs = k * x

where k is the force constant of the spring and x is the maximum distance the spring is stretched without moving the mass.

User Farbiondriven
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5.0k points
2 votes

Answer:


x=0.0049\ m= 4.9\ mm


d=0.01153\ m=11.53\ mm

Step-by-step explanation:

Given:

  • mass of the object,
    m=0.75\ kg
  • elastic constant of the connected spring,
    k=150\ N.m^(-1)
  • coefficient of static friction between the object and the surface,
    \mu_s=0.1

(a)

Let x be the maximum distance of stretch without moving the mass.

The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.


f=k.x


\mu_s.N=k.x

where:

N = m.g = the normal reaction force acting on the body under steady state.


0.1* (9.8* 0.75)=150* x


x=0.0049\ m= 4.9\ mm

(b)

Now, according to the question:

  • Amplitude of oscillation,
    A= 0.0098\ m
  • coefficient of kinetic friction between the object and the surface,
    \mu_k=0.085

Let d be the total distance the object travels before stopping.

Now, the energy stored in the spring due to vibration of amplitude:


U=(1)/(2) k.A^2

This energy will be equal to the work done by the kinetic friction to stop it.


U=F_k.d


(1)/(2) k.A^2=\mu_k.N.d


0.5* 150* 0.0098^2=0.0850 * 0.75* 9.8* d


d=0.01153\ m=11.53\ mm

is the total distance does it travel before stopping.

User Greyson
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5.3k points