what is the work to be done to increase the velocity from 18 km per hour to 19 km per hour from the mass of care is 2000...
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Given:
Initial velocity u=18km/h=5m/su=18km/h=5m/s
Final velocity v=19km/h=5.278m/sv=19km/h=5.278m/s
Mass of the car m=2000kgm=2000kg
To find:
Work done in increasing the velocity of the car from uuto v.v.
Solution:
Let WWbe the work and ΔK=Kf−KiΔK=Kf−Kibe the increase in kinetic energy. (Here KfKfand KiKiare the final and initial kinetic energies respectively.)
From the work-energy theorem, we know that the work done will equal the change in kinetic energy. That is W=ΔK.W=ΔK.
Kinetic energy of a particle K=12mv2K=12mv2
Hence, the increase in kinetic energy can be found by subtracting the initial kinetic energy from the final kinetic energy
ΔK=12mv2−12mu2ΔK=12mv2−12mu2
=> 12m(v2−u2)12m(v2−u2)
= 12(2000)(5.2782−52)12(2000)(5.2782−52)
= 5714.56825714.5682
= 2854.28J.2854.28J.
Hence the work done in increasing the velocity of a car of mass 2000 kg from 18 km/h to 19 km/h is 2854.28J.2854.28J.
And that completes the solution.
Given:
Initial velocity u=18km/h=5m/su=18km/h=5m/s
Final velocity v=19km/h=5.278m/sv=19km/h=5.278m/s
Mass of the car m=2000kgm=2000kg
To find:
Work done in increasing the velocity of the car from uuto v.v.
Solution:
Let WWbe the work and ΔK=Kf−KiΔK=Kf−Kibe the increase in kinetic energy. (Here KfKfand KiKiare the final and initial kinetic energies respectively.)
From the work-energy theorem, we know that the work done will equal the change in kinetic energy. That is W=ΔK.W=ΔK.
Kinetic energy of a particle K=12mv2K=12mv2
Hence, the increase in kinetic energy can be found by subtracting the initial kinetic energy from the final kinetic energy
ΔK=12mv2−12mu2ΔK=12mv2−12mu2
=> 12m(v2−u2)12m(v2−u2)
= 12(2000)(5.2782−52)12(2000)(5.2782−52)
= 5714.56825714.5682
= 2854.28J.2854.28J.
Hence the work done in increasing the velocity of a car of mass 2000 kg from 18 km/h to 19 km/h is 2854.28J.2854.28J.
And that completes the solution.
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