Let's solve the problem step by step. --- ## **Step 1: Draw the Free-Body Diagram Point A** At point A (just before the curve, on a straight/horizontal track): - **Gravity** acts downward: \( mg \) - **Normal force** acts upward: \( N \) - **Friction force** (kinetic, since the car is sliding): \( f_k = \mu_k N \) - The **deceleration** is caused by friction. --- ## **Step 2: Find the Normal Force at A** Since the track at A is horizontal and straight, the normal force balances gravity: \[ N = mg \] --- ## **Step 3: Friction Force** The friction force is: \[ f_k = \mu_k N = \mu_k mg \] --- ## **Step 4: Newton’s Second Law (Deceleration)** Friction is the only horizontal force slowing the car, so: \[ f_k = ma \] \[ \mu_k mg = ma \] \[ a = \mu_k g \] where: - \( a \) = deceleration - \( \mu_k \) = coefficient of kinetic friction = .20 - \( g \) = acceleration due to gravity = \( 9.81 \, \text{m/s}^2 \) --- ## **Step 5: Plug in the Values** \[ a = .20 \times 9.81 \, \text{m/s}^2 \] \[ a = 1.962 \, \text{m/s}^2 \] --- ## **Step 6: Final Answer** ### **The initial deceleration of the car as it almost reaches A is:** \[ \boxed{1.962 \ \text{m/s}^2} \] --- **Explanation:** - The deceleration is just due to friction. - On a horizontal track, normal force equals the weight. - The frictional force is \(\mu_k mg\), so deceleration is \(\mu_k g\).