Let's break down the problem: **Given:** - Mass of cabinet, \( m = 5 \) kg - Initial velocity, \( u = 5 \) m/s - Impulse force, \( F = 10 \) N - Final velocity, \( v = 0 \) m/s (since the cabinet stops) - Neglect friction We need to calculate the time \( t \) it takes the cabinet to stop. ### Step 1: Use Newton's Second Law The force applied causes a change in momentum: \[ F = \frac{\Delta p}{\Delta t} \] where \( \Delta p \) is the change in momentum. ### Step 2: Calculate Change in Momentum \[ \Delta p = m(v-u) = 5 \times (0 - 5) = -25 \text{ kg m/s} \] ### Step 3: Plug Into the Formula \[ F = \frac{\Delta p}{t} \] \[ 10 = \frac{25}{t} \] (Negative sign indicates direction, but we only need magnitude for time) ### Step 4: Solve for \( t \) \[ t = \frac{25}{10} \] \[ t = 2.5 \text{ s} \] **The correct answer is:** \[ \boxed{2.5 \text{ s}} \] So, select **2.5 s** as your answer.