Question

during a road test, a driver brakes a 1420 kg car moving at 64.8 km/h w. the car slows down and comes to a stop after moving 729 m w.
(a) calculate the net force acting on the car.
(b) what is the force of friction acting on the car while it is slowing down?

Explanation:

Part (a)

Step 1: Convert speed to m/s

The initial speed \( v_i = 64.8 \, \text{km/h} \). To convert to m/s, use \( 1 \, \text{km} = 1000 \, \text{m} \) and \( 1 \, \text{h} = 3600 \, \text{s} \). So \( v_i=\frac{64.8\times1000}{3600}= 18 \, \text{m/s} \). The final speed \( v_f = 0 \, \text{m/s} \), displacement \( d = 729 \, \text{m} \), mass \( m = 1420 \, \text{kg} \).

Step 2: Find acceleration using kinematic equation

Use the kinematic equation \( v_f^2=v_i^2 + 2ad \). Rearranging for \( a \): \( a=\frac{v_f^2 - v_i^2}{2d} \). Substitute values: \( a=\frac{0 - 18^2}{2\times729}=\frac{- 324}{1458}\approx - 0.222\,\text{m/s}^2 \) (negative sign indicates deceleration).

Step 3: Calculate net force using Newton's second law

Newton's second law: \( F_{net}=ma \). Substitute \( m = 1420 \, \text{kg} \) and \( a=- 0.222\,\text{m/s}^2 \): \( F_{net}=1420\times(- 0.222)\approx - 315.24\,\text{N} \). The negative sign indicates direction opposite to motion (east), but magnitude is approximately \( 315 \, \text{N} \) (or more precisely, let's recalculate \( a \) exactly: \( a=\frac{- 324}{1458}=-\frac{1}{4.5}=-\frac{2}{9}\approx - 0.2222\,\text{m/s}^2 \), then \( F_{net}=1420\times(-\frac{2}{9})=\frac{- 2840}{9}\approx - 315.56\,\text{N} \))

When the car is slowing down, the net force is provided by the frictional force (assuming no other horizontal forces). From Newton's second law and the fact that the net force in the horizontal direction is due to friction, the frictional force \( F_f \) has the same magnitude as the net force (since it's the only horizontal force causing deceleration) and opposite direction to motion. So \( F_f = |F_{net}| \) (in magnitude, direction is [E]).

Answer:

(a):
The net force acting on the car is approximately \(\boldsymbol{- 316 \, \text{N}}\) (or \(\boldsymbol{316 \, \text{N}}\) [E], magnitude \(\boldsymbol{316 \, \text{N}}\))

Part (b)