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a)$-7.35 \textrm{m/s}$
b)$1.29 \times 10^3 \textrm{N}$

Giancoli 6th Edition, Chapter 4, Problem 17

(2:26)

Chapter 4, Problem 17 is solved.

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Transcript for this Giancoli solution

We’ll let this box represent the two parachuters and they’re experiencing a force of air resistance upwards of zero point two five times the force of gravity, one quarter of their weight in other words, so that’s zero point two five times ‘m’ times ‘g’ and then they have the force of gravity ‘Fg’ downwards, ‘m’ times ‘g’. In part a we’ll figure out the acceleration of these parachuters, so we have the up force ‘Fa’ which is the air resistance minus down force ‘Fg’ equals ‘m’ times ‘a’. Divide both sides by ‘m’ we have acceleration is air resistance minus gravity divided by mass so that’s zero point two five times ‘m’ times ‘g’ minus ‘m’ times ‘g’ divided by ‘m’ and ‘m’ is a common factor in the numerator and denominator and so it cancels leaving us with acceleration is zero point two five ‘g’ minus ‘g’ which is negative zero point seven five ‘g’ giving us negative seven point three five meters per second squared after you substitute nine point eight for ‘g’. And that’s negative so it’s downwards and that makes sense because the parachuters are falling faster and faster. In part b we have to answer what is the force of air resistance knowing that the velocity is constant. Since there is no acceleration it means the net force is zero and that tells that the air resistance up equals the gravity downwards which is ‘m’ times ‘g’ and so that’s one hundred and thirty two kilograms times nine point eight newtons per kilogram giving us an air resistance force of one point two nine times ten raised to power three newtons which is positive and upwards. In the 5th Edition this is one hundred and twenty kilograms instead and the 5th Edition answer for the air resistance there is one point one eight times ten raised to power three newtons upwards.