Question:
Find the average frictional force needed to stop a car weighing $500 \mathrm{~kg}$ in a distance of $25 \mathrm{~m}$ if the initial speed is $72 \mathrm{~km} / \mathrm{h}$.
Solution:
$v^{2}-u^{2}=-2 a s$
$a=\frac{\left(v^{2}-u^{2}\right)}{-2 s}$
$a=\frac{(0-400)}{-2} \times 25=8 \mathrm{~m} / \mathrm{s}^{2}$
frictional force $F=m a=500 \times 8$
$F=4000 \mathrm{~N}$
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