In the figure below m1 = 1 kg and m2 = 2 kg. These two masses are pushed together on a horizontal surface to

In the figure below m1 = 1 kg and m2 = 2 kg. These two masses are pushed together on a horizontal surface to an initial separation of xi = 5 cm. The spring that is attached to the mass m1 has a spring constant of k = 3750 N/m and an equilibrium length of 7 cm. Thus, the spring is initially compressed between the two masses. When the masses are released, they move apart from each other, but eventually come to a rest because of the friction between the masses and the horizontal surface. The coefficient of kinetic friction between the masses and the surface is μk = 0.051. What is the final separation distance xf (in m)

Given:

M1 = 1 kg
M2 = 2 kg
xi = 0.05 m
xequil = 0.07 m
k = 3750N/m
μk = 0.051

Find: The final distance (xf) between M1 and M2

Fspr = kΔx

Equation 1) Fspr = k(xequil - xi)

Equation 2) Ffric1 = μN1 = μM1g

Equation 3) Ffric2 = μN2 = μM2g

Assumption: 1/3 of the spring force pushes M1 to left and 2/3 of spring force pushes M2 to right

Sum of all forces in x direction for M1:

Equation 4) (1/3)Fspr - Ffric1 = -M1a1

(1/3)k(xequil - xi) - μM1g = -M1a1

a1 = [μM1g - (1/3)k(xequil - xi)]/M1

a1 = [0.051*1*9.81 -(1/3)*3750*(0.07 - 0.05)]/1

a1 = -24.49969m/s^2

Sum of all forces in x direction for M2:

Equation 5) (2/3)Fspr - Ffric2 = M2a2

a2 = [(2/3)Fspr - Ffric2]/M2

a2 = [(2/3)k(xequil - xi) - μM2g]/M2

a2 = [(2/3)*3750*(0.07-0.05) - 0.051*2*9.81]/2

a2 = 24.49969m/s^2

∴ a2 = -a1, meaning that M1 travels travels twice as far as M2 from the spring

TotalWork = SpringWork - FrictionalWork

M1a1d1 = (1/2)kx^2 - μM1gd1

M1a1d1 + μM1gd1 = (1/2)kx^2

d1(M1a1 + μM1g) = (1/2)kx^2

d1 = [(1/2)kx^2]/(M1a1 + μM1g)

d1 = [(1/2)*3750*(0.07-0.05)^2]/(1*-24.5+0.05…

d1 = -0.0312504037 = -0.0313 m


d2 = [(1/2)kx^2]/(M2a2 + μM2g)

d2 = [(1/2)*3750*(0.07-0.05)^2]/(2*24.5+0.051…

d2 = 0.014999814 = 0.0150 m

∴ Final distance = | d1 | + | d2 | = 0.0313 + 0.015 = 0.0463m