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A 70 kg car passenger is brought to a halt by a seat belt in 0.26 s during a crash from an initial speed of 10 m/s. What average force does the seat belt exert on the passenger?
AAbout 2700 N
BAbout 270 N
CAbout 27,000 N
DAbout 182 N
Answer & Solution
Correct answer: A. About 2700 N
1. The passenger's change in momentum is delta p equal to (70 kg)(0 m/s minus 10 m/s), with a magnitude of 700 kg m/s.
2. The average force is F_net equal to delta p divided by delta t, using the given stopping time of 0.26 s.
3. Dividing gives 700 kg m/s divided by 0.26 s, which is about 2692 N, or about 2700 N to two significant figures.
4. 270 N is off by a factor of ten, and 27,000 N is off by a factor of ten in the other direction.
5. 182 N would come from multiplying mass by time rather than dividing momentum change by time.
_Source: OpenStax College Physics (CC BY 4.0), Ch 8 "Linear Momentum and Collisions", section 8.2 Impulse_
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