A light spring with force constant 3.00 N/m is compressed by 6.00 cm as it is held between a 0.350 kg block on the left and a 0.700 kg block on the right, both resting on a horizontal surface. The spring exerts a force on each block, tending to push them apart. The blocks are simultaneously released from rest. Find the acceleration with which each block starts to move, given that the coefficient of kinetic friction between each block and the surface is the following values. Let the coordinate system be positive to the right and negative to the left.

(a) µ = 0
heavier block __m/s2
lighter block __m/s2

(b) µ = 0.035
heavier block __m/s2
lighter block __m/s2

(c) µ = 0.462
heavier block __m/s2
lighter block __m/s2

the force on each side of the spring is the same.

force=kx

Then, opposing that force is fricion on each side.

to the left, net force=kx-mu*.350*g
to the right, net force=kx-mu*.700*g

on the left= kx-mu*.350*9.8=.350*a where kx = 3*.06=.18N
on the right, kx-mu*.7*9.8=.7a

Remember, when friction is greater than kx, motion does not occur.

To find the acceleration with which each block starts to move, we need to apply Newton's second law for each block separately.

First, let's calculate the force exerted by the spring on each block. The force exerted by a spring is given by Hooke's Law, F = k * x, where F is the force, k is the force constant of the spring, and x is the displacement of the spring from its equilibrium position.

Given k = 3.00 N/m and x = 0.06 m (converted from 6.00 cm), we can calculate the force exerted by the spring on each block:

Force exerted by the spring on the heavier block:
F_heavier = k * x = 3.00 N/m * 0.06 m = 0.18 N

Force exerted by the spring on the lighter block:
F_lighter = k * x = 3.00 N/m * 0.06 m = 0.18 N

Now let's consider the forces acting on each block and apply Newton's second law:

For the heavier block:
The forces acting on the heavier block are:
- Force exerted by the spring (pushing to the right) = 0.18 N
- Force of kinetic friction = µ * mass_heavier * acceleration_heavier
- Weight (downward) = mass_heavier * gravity

Since the block is initially at rest, the net force on the heavier block is zero. Therefore:

0.18 N - µ * mass_heavier * acceleration_heavier - mass_heavier * gravity = 0

Rearranging the equation, we can solve for the acceleration_heavier:

µ * mass_heavier * acceleration_heavier = 0.18 N - mass_heavier * gravity

Acceleration_heavier = (0.18 N - mass_heavier * gravity) / (µ * mass_heavier)

For the lighter block:
The forces acting on the lighter block are:
- Force exerted by the spring (pushing to the left) = 0.18 N
- Force of kinetic friction = µ * mass_lighter * acceleration_lighter
- Weight (downward) = mass_lighter * gravity

Since the block is initially at rest, the net force on the lighter block is zero. Therefore:

0.18 N - µ * mass_lighter * acceleration_lighter - mass_lighter * gravity = 0

Rearranging the equation, we can solve for the acceleration_lighter:

µ * mass_lighter * acceleration_lighter = 0.18 N - mass_lighter * gravity

Acceleration_lighter = (0.18 N - mass_lighter * gravity) / (µ * mass_lighter)

Now, we can substitute the given values of µ and the masses of the blocks to find the accelerations for each case.

(a) µ = 0:
For this case, the friction is not present since µ = 0. Therefore:

Acceleration_heavier = (0.18 N - mass_heavier * gravity) / (0 * mass_heavier)
Acceleration_lighter = (0.18 N - mass_lighter * gravity) / (0 * mass_lighter)

As dividing by zero is undefined, in the case where µ = 0, the acceleration of both blocks will be undefined.

(b) µ = 0.035:
For this case:

Acceleration_heavier = (0.18 N - mass_heavier * gravity) / (0.035 * mass_heavier)
Acceleration_lighter = (0.18 N - mass_lighter * gravity) / (0.035 * mass_lighter)

Substitute the masses of the blocks to calculate the accelerations.

(c) µ = 0.462:
For this case:

Acceleration_heavier = (0.18 N - mass_heavier * gravity) / (0.462 * mass_heavier)
Acceleration_lighter = (0.18 N - mass_lighter * gravity) / (0.462 * mass_lighter)

Substitute the masses of the blocks to calculate the accelerations for this case as well.