Question35. The driver in the previous problem applies the brakes when the car is moving at $90.0 \mathrm{~km} / \mathrm{h}$ , and the car comes to rest after traveling 40.0 m . What is the net force on the car during its deceleration?
36. An $80.0-\mathrm{kg}$ passenger in an SUV traveling at $1.00 \times 10^{2} \mathrm{~km} / \mathrm{h}$ is wearing a seat belt. The driver slams on the brakes and the SUV stops in 45.0 m . Find th force of the seat belt on the passenger.

Studdy Solution

STEP 1

1. The car in problem 35 decelerates from an initial velocity of $90.0 \, \text{km/h}$ to rest over a distance of $40.0 \, \text{m}$ .
2. The passenger in problem 36 has a mass of $80.0 \, \text{kg}$ and decelerates from an initial velocity of $100.0 \, \text{km/h}$ to rest over a distance of $45.0 \, \text{m}$ .
3. We assume constant deceleration in both problems.
4. We need to find the net force on the car in problem 35 and the force of the seat belt on the passenger in problem 36.

STEP 2

1. Convert velocities from km/h to m/s.
2. Use kinematic equations to find acceleration.
3. Use Newton's second law to find the net force.

STEP 3

Convert velocities from km/h to m/s.
For problem 35: $90.0 \, \text{km/h} = \frac{90.0 \times 1000}{3600} \, \text{m/s} = 25.0 \, \text{m/s}$
For problem 36: $100.0 \, \text{km/h} = \frac{100.0 \times 1000}{3600} \, \text{m/s} = 27.78 \, \text{m/s}$

STEP 4

Use kinematic equations to find acceleration.
For problem 35, use the equation: $v^2 = u^2 + 2as$ where $v = 0 \, \text{m/s}$ , $u = 25.0 \, \text{m/s}$ , and $s = 40.0 \, \text{m}$ .
$0 = (25.0)^2 + 2a(40.0)$ $0 = 625 + 80a$ $a = -\frac{625}{80}$ $a = -7.81 \, \text{m/s}^2$
For problem 36, use the same equation: $v^2 = u^2 + 2as$ where $v = 0 \, \text{m/s}$ , $u = 27.78 \, \text{m/s}$ , and $s = 45.0 \, \text{m}$ .
$0 = (27.78)^2 + 2a(45.0)$ $0 = 771.84 + 90a$ $a = -\frac{771.84}{90}$ $a = -8.576 \, \text{m/s}^2$

STEP 5

Use Newton's second law to find the net force.
For problem 35: Let $m$ be the mass of the car. The net force $F$ is given by: $F = ma$ Since the mass $m$ is not provided, the net force is expressed in terms of $m$ : $F = m \times (-7.81) \, \text{N}$
For problem 36: The force of the seat belt $F$ on the passenger is given by: $F = ma$ where $m = 80.0 \, \text{kg}$ and $a = -8.576 \, \text{m/s}^2$ .
$F = 80.0 \times (-8.576)$ $F = -686.08 \, \text{N}$
The force of the seat belt on the passenger is $\boxed{-686.08 \, \text{N}}$ .

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