Question

A car having a mass of 1000 Kg is moving with a speed of 25 m/s applied brake and
came to rest in 6 seconds. Which theorem can be used to find the work done to stop the
car?

Answer 1

Answer:

work done = - 3,12,500 J

distance = 75.12 m

Explanation:

mass = 1000 Kg

velocity = 25 m/s

W = work done

W = 1/2 m ( v² - u² )

W = 1/2 (1000) ( 0² - 25² )

W = 500 ( 0 - 625 )

W = 500 × - 625

W = - 3,12,500 J

finding the distance the car will take :

2as = v² - u²

acceleration = v-u /t   $\frac{v-u}{t}$

acceleration = 0 - 25 / 6

acceleration = $\frac{-25}{6}$

= -4.16 m/s²

2(-4.16) × s  = 0 - 25²

- 8.32 × s = - 625

s =  $\frac{-625}{-8.23}$

s =  $\frac{625}{8.32}$

s = 75.12 m