Physics, asked by nikhilprasad98, 6 months ago

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

Answers

Answered by rekhakumari36967
6

The force is given as,

F=ma

=500×8

=4000N

Answered by prince5132
15

\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}}}

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