**Answer:**

a) μ = 0.0136, b) F = 22.8 N

**Explanation:**

This exercise must be solved in parts. Let's start by using conservation of moment.

a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved

initial instant. Before the crash

p₀ = m v₀

final instant. After inelastic shock

p_f = (m + M) v

the moment is preserved

p₀ = p_f

m v₀ = (m + M) v

v =

We look for the speed of the block with the bullet inside

v =

v = 0.448 m / s

Now we use the relationship between work and kinetic energy for the block with the bullet

in this journey the force that acts is the friction

W = ΔK

W = ½ (m + M) - ½ (m + M) v₀²

the final speed of the block is zero

the work between the friction force and the displacement is negative, because the friction always opposes the displacement

W = - fr x

we substitute

- fr x = 0 - ½ (m + M) vo²

fr = ½ (m + M) v₀² / x

the friction force is

fr = μ N

μ = fr / N

equilibrium condition

N - W = 0

N = W

N = (m + M) g

we substitute

μ = ½ v₀² / x g

we calculate

μ = ½ 0.448 ^ 2 / 0.75 9.8

μ = 0.0136

b) Let's use the relationship between work and the variation of the kinetic energy of the block

W = ΔK

initial block velocity is zero vo = 0

F x₁ = ½ M v² - 0

F =

F = ½ 2.50 0.448² / 0.0110

F = 22.8 N