A railway engine mass 10 power 4 is moving with a speed of 72 km/h. The force which should be applied to bring it to rest over a distance of 20 m is

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Solution

Dear Student ,
here the initial speed of the train is 72 km/h and final velocity is zero .
$N o w f r o m e q u a t i o n o f m o t i o n w e c a n w r i t e t h a t, v^{2} - u^{2} = 2 f s \Rightarrow f = \frac{v^{2} - u^{2}}{2 s} \Rightarrow f = \frac{0 - {(72 \times \frac{5}{18})}^{2}}{2 \times 20} \Rightarrow f = - 10 m / s^{2} N e g a t i v e s i g n m e a n s t h a t t h e a c c e l e r a t i o n i s n e g a t i v e t h a t i s r e t a r d a t i o n m o t i o n . N o w, f o r c e r e q u i r e d t o s t o p t h e e n g i n e b y a d i s \tan c e 20 m i s, F = m f = 10^{4} \times 10 = 10^{5} N$
Regards

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