Question

Find the average force needed to accelerate a car weighing 500kg from rest to 72 km/h in a distance of 25 m

Answer 1

Explanation:

A = (v^2 - u^2) / (2S)

= [(72 × 5/18 m/s)^2 - 0] / (2 × 25 m)

= 400 / 50 m

= 8 m/s^2

F = m a

= 500 kg × 8 m/s^2

= 4000 N

= 4 kN

Answer 2

Explanation:

$\pink{ \frak{Given}} \begin{cases} \sf \: initial \: velocity \: (u) = 0 \: ms ^{ - 1} \\ \\ \sf \: final \: velcity \: (v) = 20 \: ms ^{ - 1} \\ \\ \sf \: mass\: (m) = 500 \: kg \\ \\ \sf \: distance \: (s) = 25 \: m\end{cases}$

☯$\underline{\boldsymbol{According\: to \:the\: Question :}}$

As we know that,

$\\ : \implies \displaystyle \sf \: v ^{2} - u ^{2} = 2as$

$: \implies \displaystyle \sf \:(20) ^{2} - (0) ^{2} = 2 \times a \times 25$

$: \implies \displaystyle \sf \:400 - 0 = 2 \times 25a$

$: \implies \displaystyle \sf \:400 = 2 \times 25a$

$: \implies \displaystyle \sf \: \frac{400}{2} = 25a$

$: \implies \displaystyle \sf \:a = \frac{200}{25}$

$: \implies \underline{ \boxed{\displaystyle \sf \bold{\:a = 8 \: ms ^{ - 2} }} }$

_____________________…

$\\ \dashrightarrow \displaystyle \sf \: Force = mass \times Acceleration$

$\dashrightarrow \displaystyle \sf \: Force = 500 \: kg \times 8 \: ms ^{ - 2}$

$\dashrightarrow \displaystyle \sf \: Force = 4000 \: kg.ms ^{ - 2}$

$\dashrightarrow \underline{ \boxed{\displaystyle \sf \: \bold{ Force = 4000 \: N}}}$