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A driver, traveling at 22.0 m/s, slows down her car and stops. What work is done by the friction force against the wheels if the mass of the car is 1500 kg? Group of answer choices -3.6 x 105 joules None of the above 3.6 x 106 joules -4.5 x 106 joules 4.5 x 105 joules

Respuesta :

barmansuraj489

-3.6× 10^5J is the Work done by friction force

Work done = force × distance

v^2 = u^ +2as

u= 22m/s

a =10m/s^2

When the car stops the final velocity (v) =0

0= 22^2 +2×10×s

s = -484/20

s =-24.2m

Work done = force × distance

Force = mass × acceleration

Work done = 1500×10× -24.2

= -3.6×10^5J

Hence the Work done is -3.6×10^5J

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