Question

what will be the work required to stop a car of mass 1200 kg moving with a velocity of 60 km/h

Answer 1

Mass=1200 kg
Intial Velocity (u)=60 km/h=50/3 m/s
Final Velocity (v)=0 km/h=0 m/s

Change Kinetic energy
=1/2mv²-1/2mu²
=1/2×1200×0²-1/2×1200×(50/3)²
=600×0-(600×50/3×50/3)
=0-(166666.667)
=-166666.66 J

Work=-166666.66J

Answer 2

DATA GIVEN IN THE QUESTION :

➡️Mass=1200kg
➡️Initially the velocity of car is (u)=60km/h=60×1000/3600=60000/3600=50/3m/sec
➡️Final velocity of car (v)=0

⏺️Calculation:

⏩WE KNOW THAT:
CHANGE IN KINETIC ENERGY=
$\frac{1}{2} m {v}^{2} - \frac{1}{2} m {u}^{2}$

$\frac{1}{2} m( {v}^{2} - {u}^{2} ) \\ \\ According \:to\: question\\ \frac{1}{2} \times 1200( {0}^{2} - \frac{50}{3} \times \frac{50}{3} ) \\ \\ 600( \frac{-2500}{9} ) \\ work done = -166666.6666 joules$

(here -sign shows negative work done )

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