Question

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.

Answer 1

Answer:

$x=0.0049\ m= 4.9\ mm$

$d=0.01153\ m=11.53\ mm$

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.

$0.1\times (9.8\times 0.75)=150\times x$

$x=0.0049\ m= 4.9\ mm$

(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:

$U=\frac{1}{2} k.A^2$

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

$U=F_k.d$

$\frac{1}{2} k.A^2=\mu_k.N.d$

$0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d$

$d=0.01153\ m=11.53\ mm$

is the total distance does it travel before stopping.