please help An air-track glider of mass 0.098 kg is attached to the end of a hor

Question

please help

An air-track glider of mass 0.098 kg is attached to the end of a horizontal air track by a spring with force constant 16.5 N/m With the air track turned off, the glider travels 8.9 cm before it stops instantaneously. How large would the coefficient of static friction mus have to be to keep the glider from springing back to the left? If the coefficient of static friction between the glider and the track is mus= 0.57, what is the maximum initial speed v1 that the glider can be given and remain at rest after it stops instantaneously? With the air track turned off, the coefficient of kinetic friction is muk = 0.54. m/s

Explanation / Answer

a.)F=kx=umg

16.5*0.089=u*.098*9.8

So u=1.53

b.)the KE of the glider

.5*m*v^2
will be converted to potential energy in the spring
.5*k*x^2
and friction
m*g*uk*x

set up the energy equation

.5*m*v^2=.5*k*x^2+m*g*uk*x
==================
the second condition is that
k*x<=m*g*us

or as the max
k*x=m*g*us
or
x=m*g*us/k

plug into the energy equation and solve for v
==================
.5*m*v^2=
.5*m^2*g^2*us^2/k+m^2*g^2*uk*us/k


v^2=m*g^2*us*(us+uk)/k

v=sqrt(m*g^2*us*(us+uk)/k)
plug in the numbers

vmax=0.6 m/s   

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