on a level road a scooterist applies brakes to slow down from a speed of 10m/s to 5m/s. If the mass of the scooter and the scooterist be 150kg ,calculate the work done by the brakes. (neglect air resistance and friction).
Answers
Answered by
261
Work done = Change in Kinetic energy
= 0.5mv² - 0.5mu²
= 0.5m (v² - u²)
= 0.5 × 150 kg × [(5 m/s)² - (10 m/s)²]
= -5625 J
= -5.625 kJ
Work done by brakes is -5.625 kJ
Here negative sign indicates that work is done against the motion of scooter.
= 0.5mv² - 0.5mu²
= 0.5m (v² - u²)
= 0.5 × 150 kg × [(5 m/s)² - (10 m/s)²]
= -5625 J
= -5.625 kJ
Work done by brakes is -5.625 kJ
Here negative sign indicates that work is done against the motion of scooter.
JunaidMirza:
You're welcome
Answered by
46
Answer:
-5625.0 or -5625 j
Explanation:
As we know that,
Work done = Change in kinetic energy
0.5mv^2 = 0.5mu^2
0.5m(v^2-u^2)
0.5*150*-75
37.5*-150
-5625 joules
- ve sign indicates that work is acting on the opposite side of force.
Hope it will help you.
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