You push a 1kg block against a spring with spring constant k = 100 Nmon a horizontal table top. The spring is compressed to 0.2m away from its relaxed length, and then the block is released. If the horizontal table top has a

coefficient of kinetic friction of 0.2, how much further past the end of the spring does the block go before coming to rest?

energy stored in spring = (1/2) k x^2

= (1/2)(100)(.04)

work done by friction
= mu m g *d

so
50*.04 = .2 * 1 *9.81* d

solve for d

To determine how much further past the end of the spring the block goes before coming to rest, we need to analyze the forces acting on the block.

1. Initially, the block is pressed against the spring and is not moving. The only force acting on the block is the force exerted by the spring, which follows Hooke's Law: F_spring = -k * x, where k is the spring constant and x is the displacement from the relaxed position.

Given that the spring constant k = 100 N/m and the spring is compressed to 0.2 m, the spring force can be calculated as F_spring = -100 N/m * 0.2 m = -20 N (negative sign indicates that the force is acting in the opposite direction to the displacement).

2. When the block is released, it experiences the force of friction from the table. The force of friction can be calculated using the formula F_friction = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.

The normal force is equal to the weight of the block, which can be calculated as N = m * g, where m is the mass of the block and g is the acceleration due to gravity. Given that the mass of the block is 1 kg and g is approximately 9.8 m/s², the normal force is N = 1 kg * 9.8 m/s² = 9.8 N.

Therefore, the force of friction can be calculated as F_friction = 0.2 * 9.8 N = 1.96 N.

3. The net force acting on the block is the sum of the spring force and the force of friction, since they act in opposite directions: F_net = F_spring + F_friction.

Substituting the values, we have F_net = -20 N + 1.96 N = -18.04 N.

4. The acceleration of the block can be determined using Newton's second law, which states that F_net = m * a, where F_net is the net force and m is the mass of the object.

Rearranging the equation, we get a = F_net / m = -18.04 N / 1 kg = -18.04 m/s².

Note: The negative sign indicates that the block is experiencing deceleration.

5. To find how much further the block moves before coming to rest, we use the equations of motion for uniformly decelerated motion. In this case, the equation to use is v² = u² - 2 * a * s, where v is the final velocity (which is 0 m/s since the block comes to rest), u is the initial velocity (which is 0 m/s since the block starts from rest), a is the deceleration, and s is the distance traveled.

Substituting the values, we have 0 = 0 - 2 * (-18.04 m/s²) * s.

Simplifying the equation, we get s = 0.5 * (-18.04 m/s²)^2.

Calculating the value, we find s ≈ 16.23 m.

Therefore, the block moves approximately 16.23 meters further past the end of the spring before coming to rest.