what is the work to be done to increase the velocity of a car from 30 kilometre h-¹ to 60 km h -¹ if the mass of the car is 1500kg?
Answers
Answer:
Answer:
156375 J
Solution:-
Mass of the car=1500 kg
Intial velocity of car=30 km h^-1
= \frac{30 \times 1000 \: m}{60 \times 60 \: sec}=
60×60sec
30×1000m
= 8.33 \: m \: {s}^{ - 1}=8.33ms
−1
Similarly;the final velocity of the car
v = 60 \: km \: {s}^{ - 1}v=60kms
−1
= 16.67 \: m \: {s}^{ - 1}=16.67ms
−1
The Intial kinetic energy of the car
e_{ki} = \frac{1}{2}m \: {u}^{2}e
ki
=
2
1
mu
2
= \frac{1}{2} \times 1500 \: kg \times {(8.33m \: {s}^{ - 1} }^{2})=
2
1
×1500kg×(8.33ms
−1
2
)
= 52041.68 \: j=52041.68j
The final kinetic energy of the car
e_{kf} = \frac{1}{2} \times \frac{1}{2} \times {(16.67m \: {s}^{ - 1} })^{2}e
kf
=
2
1
×
2
1
×(16.67ms
−1
)
2
= 208416.68 \: j=208416.68j
_____________________
Thus the work done=change in kinetic energy
= e_{kf} - e_{ki}=e
kf
−e
ki
= 208416.68 - 52041.68=208416.68−52041.68
=156375 J
Explanation:
mass of the car =1500kg
intial velocity of the car = 30kg hn-1
ifrac(30times 1000/:m) (60l times 60l:sec
60*60sec
30*1000m
2.8416.68
5204168=208416.68-52041.68=156375j