a 3000 kg truck moving at speed 72km/h stops at some distance.the force applied by the brakes is 24000N.compute the distance covered and work done by this force
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Answered by
3
(72×5÷18)=20m/s
since W=1/2mv2
so,W=1/2×3000×20×20=600000J
now since W=fs
→600000=24000×s
→s=600000/24000=25m
since W=1/2mv2
so,W=1/2×3000×20×20=600000J
now since W=fs
→600000=24000×s
→s=600000/24000=25m
Answered by
2
(72×5÷18)=20m/s
since W=1/2mv2
so,W=1/2×3000×20×20=600000J
now since W=fs
→600000=24000×s
→s=600000/24000=25m
since W=1/2mv2
so,W=1/2×3000×20×20=600000J
now since W=fs
→600000=24000×s
→s=600000/24000=25m
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