A block of mass 2 kg is projected with an initial speed of 14 m/s...

The deceleration of the block due to kinetic friction is given by:

$a = -\mu_k g = -0.2 \times 10 = -2 \text{ m/s}^2$

Let $v_1$ be the velocity of the block just before it collides with the bumper. Using the third equation of motion:

$v_1^2 = u^2 + 2a d_1$

$v_1^2 = (14)^2 + 2(-2)(24)$

$v_1^2 = 196 - 96 = 100 \Rightarrow v_1 = 10 \text{ m/s}$

Let $v_2$ be the velocity of the block just after rebounding. Since it comes to rest after traveling $d_2 = 9 \text{ m}$:

$0 = v_2^2 + 2a d_2$

$v_2^2 = 2(2)(9) = 36 \Rightarrow v_2 = 6 \text{ m/s}$

The impulse $J$ imparted by the bumper is equal to the change in momentum of the block. Taking the initial direction of motion as positive:

$J = m(v_f - v_i) = m(-v_2 - v_1)$

$J = 2(-6 - 10) = -32 \text{ N s}$

The magnitude of the impulse is $32 \text{ N s}$.

Answer: $32 \text{ N s}$