Question

An air-track glider of mass 0.109 kg is attached to the end of a horizontal air track by a spring with force constant 24.5 N/m

(a) With the air track turned off, the glider travels 8.0 cm before it stops instantaneously. How large would the coefficient of static friction μs have to be to keep the glider from springing back to the left?

(b) If the coefficient of static friction between the glider and the track is μs= 0.65, 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 μk= 0.51.

Answer 1

a) Sum of the horizontal forces on the block = 0 <---- no movement
0 = F - friction ( as after stretching spring's tends to move the glider to left but then Friction acts on opposite of direction of movement)
F = friction

From Hooke's Law for spring
F = k*x .................(1)

Sum of the vertical force on the block (upwards is considered to be positive) = 0
0 = N - m*g
N = m*g

then we know that friction force is $F_k=\mu_sN$ ...........(2)

equating equation 1 and 2 according to given condition, we'll get

$F_k=\mu_sN= kx$

   $\mu_s(mg)= kx$ ...............(3)

   $\mu_s= 24.5(0.08)/(0.109 \times 9.8)= 1.83$

b). Now if the coefficient of static friction is 0.65, then the maximum distance at which is can remain at rest is

using equation 3,

$\mu_s(mg)=kx'$

$x'=\mu_s(mg)/k= (0.65)(0.109)(9.8)/24.5= 0.0283 m$

Now with this much displacement, restoring potential energy of spring must be comphensated by kinetic energy and friction energy of glider to remain at rest.

thus, ( since friction force leads to the loses)

where v is the maximum initial speed given,

then, $(1/2)kx'^2=(1/2)mv^2-\mu_k(mg) x'$

   $(1/2)(24.5)(0.0283)^2=(1/2)(0.109)v^2-(0.51)(0.109)(9.8)(0.0283)$

   $0.00981=0.0545v^2-0.0154$

$v=\sqrt{0.02522/0.0545}= 0.680m/s$ ( ANS)