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Chapter 3: Problem 6

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 force (assumed constant) acts on the passenger's upper torso, which has a mass of \(39 \mathrm{~kg}\) ?

Short Answer

Expert verified
The force acting on the passenger's upper torso is approximately \(-6512.86 \mathrm{N}\)

Step by step solution

01

Convert Units

Convert all quantities to standard units. Speed is given as \(53 \mathrm{~km} / \mathrm{h}\), which should be converted to \(\mathrm{m/s}\). This can be done by multiplying by \(\frac{1000}{3600}\). Distance \(65 \mathrm{~cm}\) should be converted to meters by dividing by 100. So the speed becomes \(14.72 \mathrm{m/s}\) and distance becomes \(0.65 \mathrm{m}\).
02

Find Acceleration

Use the equation of motion \(v^2 = u^2 + 2as\), where \(v\) is final velocity, \(u\) is initial velocity, \(a\) is acceleration, and \(s\) is distance. Here, \(v\) is \(0 \mathrm{m/s}\), \(u\) is \(14.72 \mathrm{m/s}\), and \(s\) is \(0.65 \mathrm{m}\). Rearrange the formula to solve for acceleration, \(a = \frac{v^2 - u^2}{2s}\), and substitute the known values to find \(a = \frac{(0)^2 - (14.72)^2}{2*0.65} = -166.74 \mathrm{m/s^2}\) (The negative sign indicates the direction of acceleration).
03

Find Force

Use Newton's second law, \(F = ma\), to calculate the force. Substitute \(m = 39 \mathrm{kg}\), \(a = -166.74 \mathrm{m/s^2}\) into the formula to get \(F = 39*-166.74 = -6512.86 \mathrm{N}\). Note that the force is negative, indicating it is in the opposite direction of the motion of the car.

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