8. A car weighing 1000 kg travelling along a level
road is accelerated for 2 min at the rate of
15 m st. Find its gain of kinetic energy during
those 2 min.
[Ans. 1.62 x 10' J]
Answers
Explanation:
Given,
m=1000 kga=15m/s
2t=2minutes=2×60=120 su=0Using
relation,
v=u+at=0+15×120=1800 m/s
Gain in kinetic energy=12mv2−12mu2
=12×1000(18002−0)
=1620000000 J=1620000kJ
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The gain in kinetic energy during that 2 min is 1.62 × 10^8 J.
Given: A car weighing 1000 kg traveling along a level road is accelerated for 2 min at the rate of 15 m/s².
To Find: The gain of kinetic energy during that 2 min.
Solution:
- We need to first use the formula to find the final velocity of the car.
v = u + at .....(1)
Where v = final velocity, u = initial velocity, a = acceleration, t = time.
- It is to be assumed that the car started from rest and hence initial velocity is equal to zero.
- The gain in kinetic energy of the car can be calculated using the formula,
Δ K.E = 1/2 × m × ( v² - u² ) ....(2)
Where m = mass of the car, v = final velocity, u = initial velocity.
Coming to the numerical, we are given that;
The mass of the car = 1000 kg
The acceleration of the car = 15 m/s²
The time = 2 min = ( 2 × 60 ) s
= 120 s
The initial velocity (u) = 0
Putting respective values in (1), we get;
v = u + at
⇒ v = 0 + ( 15 × 120 )
⇒ v = 1800 m/s
Now, putting the respective values in (2), we get;
Δ K.E = 1/2 × m × ( v² - u² )
⇒ Δ K.E = 1/2 × 1000 × (( 1800 )² - 0² )
⇒ Δ K.E = 1/2 × 324000000
⇒ Δ K.E = 162000000 J
= 1.62 × 10^8 J
Hence, the gain in kinetic energy during that 2 min is 1.62 × 10^8 J.
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