#
Find the force required to stop a car of mass 100Kg with two passengers of

50 g each sitting inside,if it is moving at 60km/h speed and takes 5s to stop.

It is given that:

- Mass of car = 100 kg
- Two passengers with each of a mass of 50 g = 0.05 kg
- Initial velocity = 60 km/h = 60 (5/18) = 50/3 m/s
- Car takes 5 s to stop

Initial velocity u = 50/3 m/s

Final velocity v = 0

$v=u+at\phantom{\rule{0ex}{0ex}}a=\frac{v-u}{t}=\frac{0-50/3}{5}=-\frac{10}{3}m/{s}^{2}\phantom{\rule{0ex}{0ex}}Negativesignshowsthatthisisdeceleration.\phantom{\rule{0ex}{0ex}}Totalmass=massofcar+massof2passenger=100+2(0.050)=100+0.1=100.1Kg\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Force=ma\phantom{\rule{0ex}{0ex}}Herearepresentsdeceleration\phantom{\rule{0ex}{0ex}}F=100.1\times \frac{10}{3}=\frac{1001}{3}=333.67N\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Regards

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