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In this question(see attachment) car is moving above hill.So,why for calculating work done we use mv2/2-mgh as it was moving above the hill?Please explain this question to me.
11 months ago

Here given in the question :-
Car Mass m = 1400 kg .
h = 10 m
Moving at Initial speed v= 54 km/h = (54000/3600) m/s = 15 m/sec.
Now Initial Kinetic Energy of the car ,
= 1/2 mv²
= 1/2 (1400 x 15²)
= 315000/2
= 157500 J.
Here final velocity and Final Kinetic energy both are Zero.
Now change in Kinetic Energy
= Final - Initial
= 0 - 157500
= -157500 J.
So change in Kinetic energy is also equal to W.D by all force.
Thus work done by gravity,
= mgh
= 1400 x 9.8 (-10)
= 1400 x -98
= -137200 .
Here, change in Kinetic energy = Work done by the friction + Work done by the gravity.
-157500 = Work done by friction + (- 137200)
Work done by the friction = -157500 +137200
=  -20,300
Hence, the work done against the friction =  20300 J

11 months ago
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