A truck of
mass 1000 kg increases its velocity from 60 km/h
to 80 km/h in 10 sec. Calculate the applied force of the truck
Answers
Answer:
Solution:
Weight of truck = Force = 1000 kgf
Mass = 1000 kg
Initial velocity = 36 km\ h^{-1}km h
−1
= \frac{36\times1000}{60\times60}
60×60
36×1000
= 10 m\ s^{-1}m s
−1
New velocity = 72 km\ h^{-1}km h
−1
= \frac{72\times1000}{60\times60}
60×60
72×1000
= 20 m\ s^{-1}m s
−1
Work done = Increase in energy
= \frac{1}{2}mv^2-\frac{1}{2}mu^2
2
1
mv
2
−
2
1
mu
2
= \frac{1}{2}m\left(v+u\right)\left(v-u\right)
2
1
m(v+u)(v−u)
= \frac{1}{2}\times1000\times30\times10
2
1
×1000×30×10
= 150000 J
= 1.5\times10^5\1.5×10
5
J
Explanation:
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Answer:
Solution:
Weight of truck = Force = 1000 kgf
Mass = 1000 kg
Initial velocity = 36 km\ h^{-1}km h
−1
= \frac{36\times1000}{60\times60}
60×60
36×1000
= 10 m\ s^{-1}m s
−1
New velocity = 72 km\ h^{-1}km h
−1
= \frac{72\times1000}{60\times60}
60×60
72×1000
= 20 m\ s^{-1}m s
−1
Work done = Increase in energy
= \frac{1}{2}mv^2-\frac{1}{2}mu^2
2
1
mv
2
−
2
1
mu
2
= \frac{1}{2}m\left(v+u\right)\left(v-u\right)
2
1
m(v+u)(v−u)
= \frac{1}{2}\times1000\times30\times10
2
1
×1000×30×10
= 150000 J
= 1.5\times10^5\1.5×10
5
J