Key Concept: Force of Friction, Relative Motion
c) 4000 N
[Solution Description]
First, calculate the deceleration of the car using the formula $a = \frac{v_f - v_i}{t}$, where $v_f$ is the final velocity, $v_i$ is the initial velocity, and $t$ is the time taken to stop. Here, $v_f = 0 \, \text{m/s}$, $v_i = 20 \, \text{m/s}$, and $t = 5 \, \text{s}$. So,
$a = \frac{0 - 20}{5} = -4 \, \text{m/s}^2$
The negative sign indicates deceleration. Now, calculate the frictional force using Newton's second law $F = ma$, where $m$ is the mass of the car and $a$ is the deceleration. So,
$F = 1000 \, \text{kg} \times (-4) \, \text{m/s}^2 = -4000 \, \text{N}$
The magnitude of the frictional force is 4000 N.
Your Answer is correct.
c) 4000 N
[Solution Description]
First, calculate the deceleration of the car using the formula $a = \frac{v_f - v_i}{t}$, where $v_f$ is the final velocity, $v_i$ is the initial velocity, and $t$ is the time taken to stop. Here, $v_f = 0 \, \text{m/s}$, $v_i = 20 \, \text{m/s}$, and $t = 5 \, \text{s}$. So,
$a = \frac{0 - 20}{5} = -4 \, \text{m/s}^2$
The negative sign indicates deceleration. Now, calculate the frictional force using Newton's second law $F = ma$, where $m$ is the mass of the car and $a$ is the deceleration. So,
$F = 1000 \, \text{kg} \times (-4) \, \text{m/s}^2 = -4000 \, \text{N}$
The magnitude of the frictional force is 4000 N.
