Question
A car weighing 1000kg is going up an incline with a slope of 2 in 25 at a steady speed of 18kmph.If g=10m/s2, the power of its engine is
Answers
Henry
Tan A = 2/25 = 0.08, A = 4.57o.
Fp = Mg*sin A = 10,000*sin4.57 = 797.5 N. = Force parallel to incline.
V = 18km/h = 18,000m/3600s. = 5 m/s.
P = Fp*V = 797.5 * 5 = 3988 J./s = 3988 W.
Fp = Mg*sin A = 10,000*sin4.57 = 797.5 N. = Force parallel to incline.
V = 18km/h = 18,000m/3600s. = 5 m/s.
P = Fp*V = 797.5 * 5 = 3988 J./s = 3988 W.
Karan
Correct answer is 4kw
Madhavi
Say the process
Mj
The answer should be 4kW
But u got 3988. So definitely u need to check ur answer once again
But u got 3988. So definitely u need to check ur answer once again
Viji
3988 is in watt. So it should be changed to kilowatt.
3988÷1000=3.9
=4kw
3988÷1000=3.9
=4kw
Peram.Poojitha Reddy
I want provess how the 3988 value is came