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Chapter 5: Problem 20

A car traveling at \(53 \mathrm{~km} / \mathrm{h}\) hits a bridge abutment. A passenger in the car moves forward a distance of \(65 \mathrm{~cm}\) (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of \(41 \mathrm{~kg}\) ?

Short Answer

Expert verified
The force magnitude is 6829.37 N.

Step by step solution

01

Convert Units

First, convert the car's speed from kilometers per hour to meters per second, since we're measuring distance in meters.\[53 \text{ km/h} = 53 \times \frac{1000}{3600} \text{ m/s} = 14.72 \text{ m/s}\]. The distance moved is already in centimeters, convert this to meters: \[65 \text{ cm} = 0.65 \text{ m}\].
02

Find the Deceleration

Using the kinematic equation \(v^2 = u^2 + 2as\), where \(v = 0\text{ m/s}\) (final velocity), \(u = 14.72\text{ m/s}\) (initial velocity), and \(s = 0.65\text{ m}\), solve for \(a\).\[0 = (14.72)^2 + 2a(0.65)\] Simplifying gives: \[a = \frac{-(14.72)^2}{2 \times 0.65} = -166.57 \text{ m/s}^2\].
03

Calculate the Force

Use Newton's second law \(F = ma\) to find the force. The mass \(m = 41\text{ kg}\), and the acceleration \(a = -166.57\text{ m/s}^2\). Calculate \[F = 41 \times 166.57 = 6829.37 \text{ N}\]. The negative sign indicates the force is in the opposite direction of motion, but we are asked for magnitude, so ignore the sign.

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Key Concepts

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

Kinematic Equations
Kinematic equations are essential tools in physics for analyzing motion. They allow us to describe the movement of objects using variables like velocity, acceleration, and displacement. In our problem, we used the kinematic equation: \[ v^2 = u^2 + 2as \] This equation links final velocity \(v\), initial velocity \(u\), acceleration \(a\), and distance \(s\). Here, the car was initially moving at 14.72 m/s, and it was brought to a stop, meaning the final velocity \(v\) is 0 m/s. To find the deceleration \(a\), we substituted the known values into the equation and solved for \(a\). Deceleration is just negative acceleration, indicating the speed decrease.
  • Ensure all units are consistent before substituting into equations.
  • Remember that the square of velocity involves direction; hence, use absolute values for speed calculations.
Newton's Second Law
Newton's Second Law of Motion is a fundamental principle in physics, stating that the force \(F\) acting on an object is the product of its mass \(m\) and acceleration \(a\): \[ F = ma \] In our exercise, we needed to calculate the force exerted on the passenger by understanding the sudden deceleration due to the airbag deployment. Here, the mass of the passenger's torso is 41 kg, and we've previously computed the acceleration (or deceleration, as it was in this instance) to be -166.57 m/s虏.
  • Negative acceleration indicates a reduction in speed.
  • Forces can be interpreted in terms of direction as well as magnitude. In this problem, we considered the magnitude.
Unit Conversion
Unit conversion is a crucial step in solving many physics problems. It ensures that all measurements are in compatible units, which is necessary for accurate calculations. In this exercise, we converted the speed from kilometers per hour to meters per second, because the standard unit for velocity in physics is m/s. The conversion factor of 1000/3600 was used for this transformation. Similarly, distance was converted from centimeters to meters by dividing by 100.
  • Always check the units at the problem's start and ensure they fit the equations you'll use.
  • Being consistent with units helps prevent errors in calculations.
This step might seem straightforward, but it is incredibly important as incorrect units can lead to wrong results.

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Most popular questions from this chapter

Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward force (thrust) of \(3260 \mathrm{~N}\), the craft descends at constant speed; if the engine provides only \(2200 \mathrm{~N}\), the craft accelerates downward at \(0.39 \mathrm{~m} / \mathrm{s}^{2} .\) (a) What is the weight of the landing craft in the vicinity of Callisto's surface? (b) What is the mass of the craft? (c) What is the magnitude of the free-fall acceleration near the surface of Callisto?

A certain force gives an object of mass \(m_{1}\) an acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2}\) and an object of mass \(m_{2}\) an acceleration of \(3.30\) \(\mathrm{m} / \mathrm{s}^{2} .\) What acceleration would the force give to an object of mass (a) \(m_{2}-m_{1}\) and (b) \(m_{2}+m_{1}\) ?

Three forces act on a particle that moves with unchanging velocity \(\vec{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(7 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). Two of the forces are \(\vec{F}_{1}=(2 \mathrm{~N}) \hat{\mathrm{i}}+\) \((3 \mathrm{~N}) \hat{\mathrm{j}}+(-2 \mathrm{~N}) \hat{\mathrm{k}}\) and \(\vec{F}_{2}=(-5 \mathrm{~N}) \hat{\mathrm{i}}+(8 \mathrm{~N}) \hat{\mathrm{j}}+(-2 \mathrm{~N}) \hat{\mathrm{k}}\). What is the third force?

A shot putter launches a \(7.260 \mathrm{~kg}\) shot by pushing it along a straight line of length \(1.650 \mathrm{~m}\) and at an angle of \(34.10^{\circ}\) from the horizontal, accelerating the shot to the launch speed from its initial speed of \(2.500 \mathrm{~m} / \mathrm{s}\) (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of \(2.110 \mathrm{~m}\) and at an angle of \(34.10^{\circ}\), and it lands at a horizontal distance of \(15.90 \mathrm{~m}\). What is the magnitude of the athlete's average force on the shot during the acceleration phase? (Hint: Treat the motion during the acceleration phase as though it were along a ramp at the given angle.)

A \(40 \mathrm{~kg}\) skier skis directly down a frictionless slope angled at \(10^{\circ}\) to the horizontal. Assume the skier moves in the negative direction of an \(x\) axis along the slope. A wind force with component \(F_{x}\) acts on the skier. What is \(F_{x}\) if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\), and \((\mathrm{c})\) increasing at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2} ?\)
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