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Chapter 5: Problem 36

An 80.0-kg passenger in an SUV traveling at \(1.00 \times 10^{3} \mathrm{km} / \mathrm{h}\) is wearing a seat belt. The driver slams on the brakes and the SUV stops in \(45.0 \mathrm{m}\). Find the force of the seat belt on the passenger.

Short Answer

Expert verified
The force of the seat belt on the passenger is approximately \(12298 N\).

Step by step solution

01

Convert the initial velocity to m/s

First, we need to convert the initial velocity from km/h to m/s. Given initial velocity, v = 1.00 x 10^3 km/h 1 km = 1000 m and 1 h = 3600 s So, v = (1.00 x 10^3 * 1000) / 3600 m/s
02

Find the initial acceleration using one of the equations of motion

For this step, we will use the equation of motion, \(v^2 = v_0^2 + 2ad\), where 'v' is the final velocity (0 m/s in this case, since the car came to a stop), \(v_0\) is the initial velocity, 'a' is the acceleration, and 'd' is the braking distance. 0 = \(v_0^2 + 2ad\) We have to find 'a': a = -\(\frac{v_0^2}{2d}\)
03

Calculate the initial acceleration

Now, we can plug in the values to find the initial acceleration 'a': a = -\(\frac{(1000/3.6)^2}{2 \cdot 45}\) m/s虏 Make sure to calculate the value in m/s虏.
04

Use Newton's second law to find the force of the seat belt on the passenger

Newton's second law states that F = ma (Force = mass x acceleration). Now we can find the force of the seat belt on the passenger: Force = (mass of the passenger) x (initial acceleration) Force = (80.0 kg) x (a) N
05

Calculate the force of the seat belt on the passenger

Plug in the value of 'a' from step 3 and calculate the force: Force = (80.0 kg) x (a) N The result will give you the force of the seat belt on the passenger in Newtons (N).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics in Motion
Kinematics is a branch of mechanics that deals with the motion of objects without considering the causes of this motion, such as forces. In our exercise, we're concerned with the kinematics of an SUV and its passenger as the vehicle comes to a sudden stop. To solve kinematics problems, we often use equations of motion, which relate displacement, initial and final velocities, acceleration, and time.

For instance, the equation used in the exercise, \(v^2 = v_0^2 + 2ad\), is one of the key kinematic equations, enabling us to calculate the unknown acceleration given the initial velocity and the distance traveled before stopping. Understanding these equations and knowing when and how to use them is crucial for solving kinematics-related problems in physics.
Newton's Second Law of Motion
Newton's second law is profoundly important in physics and states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass (\(F = ma\)). In the context of our problem, once the SUV begins to decelerate, the passenger continues moving forward due to inertia. The seat belt exerts a force to decelerate the passenger at the same rate as the SUV.

By applying Newton's second law, we can determine the force the seat belt applies to the passenger during this rapid deceleration period. This principle doesn't just apply to our specific problem鈥攊t is universally used to relate the motion of all objects to the forces acting upon them, and is a cornerstone of classical mechanics.
Conversion of Units
Solving physics problems accurately often requires converting between different units. In our exercise, the SUV's initial velocity was given in kilometers per hour (km/h), but to apply the kinematic equations accurately, we need to convert this velocity into meters per second (m/s), as the SI unit for velocity is m/s.

This involves multiplying the velocity value by the conversion factor stemming from the equivalences 1 km = 1000 m and 1 hour = 3600 seconds. Mastery of unit conversion is an essential skill in physics, as it ensures that the units involved in calculations are consistent, allowing us to accurately apply formulas and reach correct solutions.

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Most popular questions from this chapter

Suppose the mass of a fully loaded module in which astronauts take off from the Moon is \(1.00 \times 10^{4}\) kg. The thrust of its engines is \(3.00 \times 10^{4}\) N. (a) Calculate the module's magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If it could, calculate the magnitude of its acceleration.

A fireman has mass \(m\); he hears the fire alarm and slides down the pole with acceleration \(a\) (which is less than \(g\) in magnitude). (a) Write an equation giving the vertical force he must apply to the pole. (b) If his mass is \(90.0 \mathrm{kg}\) and he accelerates at \(5.00 \mathrm{m} / \mathrm{s}^{2},\) what is the magnitude of his applied force?

A car moves along a horizontal road. Draw a free-body diagram; be sure to include the friction of the road that opposes the forward motion of the car.

(a) Calculate the tension in a vertical strand of spider web if a spider of mass \(2.00 \times 10^{-5} \mathrm{kg}\) hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in Figure \(5.26 .\) The strand sags at an angle of \(12^{\circ}\) below the horizontal. Compare this with the tension in the vertical strand (find their ratio).

Two forces are acting on a 5.0 -kg object that moves with acceleration \(2.0 \mathrm{m} / \mathrm{s}^{2}\) in the positive \(y\) -direction. If one of the forces acts in the positive \(x\) -direction and has magnitude of \(12 \mathrm{N},\) what is the magnitude of the other force?
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