A light spring of force constant 3.85 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.500 kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. What is the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is:. A) 0
B) 0.100
C) 0.463
Assume that the coefficient of static friction is larger than that for kinetic friction.

A) v1 = -0.256 m/s

v2 = 0.128 m/s

B) v1 = -0.0642 m/s

v2 = 0 m/s

C) v1 = v2 = 0 m/s

Explanation:

We are given;

Spring constant; k = 3.85 N/m

Distance compressed; x = 8 cm = 0.08 m

Mass of left block; m1 = 0.25 kg

Mass of right block; m2 = 0.5 kg

A) From conservation of energy;

½kx² = ½m1•v1² + ½m2•(v2)²

Also from conservation of linear momentum, we know that;

m1v1 = m2v2

Now, v2 = m1•v1/m2

Plugging this for v2 in the first equation gives;

½kx² = ½m1•v1² + ½m2•(m1•v1/m2)²

Making v1 the subject, we have;

v1 = √(kx²(m2))/(m1•m2 + (m1)²))

v1 = √[(3.85 × 0.08² × 0.5)/((0.25 × 0.5) + 0.5²))

v1 = 0.256 m/s

This is the left block which is in the negative x direction and so v1 = -0.256 m/s

We saw that v2 = m1•v1/m2

Thus; v2 = (0.25 × 0.256)/0.5

v2 = 0.128 m/s

B) Force exerted by spring is;

F_s = kx = 3.85 × 0.08

F_s = 0.308 N

Normal forces will be calculated for both blocks as;

N1 = m1•g = 0.25 × 9.8 = 2.45 N

N2 = m2•g = 0.5 × 9.8 = 4.9 N

Let's calculate force of static friction for both blocks;

F_s1 = μN1 and F_s2 = μN2

We are given coefficient of friction as

μ = 0.1.

Thus;

F_s1 = 0.1 × 2.45 = 0.245 N

F_s2 = 0.1 × 4.9 = 0.49 N

F_s1 is lesser than F_s. Thus let's calculate the new compression;

x_1 = 0.245/3.85

x_1 = 0.06364 m

Thus, change in compression is;

Δx = x - x_1

Δx = 0.08 - 0.06364

Δx = 0.01636 m

From conservation of energy, since our coefficient of friction is not zero and we have frictional force, then we use the equation;

½k(x)² - F_s1•Δx = ½m1•v1² + ½k(x_1)²

Making v1 the subject, we have;

v_1 = √(k(x² - x_1²) - 2F_s1•Δx)/m1

v_1 = √(3.85(0.08² - 0.06364²) - 2(0.245 × 0.01636)/0.25

v1 = 0.0642 m/s

Since in negative x direction, then

v1 = -0.0642 m/s

F_s2 is greater than F_s. Thus, it means the right side block will not move and velocity is zero. v2 = 0 m/s

C) Coefficient of friction is now 0.463.

Thus;

F_s1 = 0.463 × 2.45 = 1.13435 N

F_s2 = 0.463 × 4.9 = 2.2687 N

They are both greater than F_s and thus no motion in both cases.

So v1 = v2 = 0 m/s

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