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Solve the following :

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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