an engine pulls a car on a level road at a constant speed of 5.0 metre per second against a frictional force of 500 Newton calculate the power extended by engine. what extra power has the engine to extend in order to maintain the same speed of the car up an inclined plane having a gradient of 1 in 10 ?
Answers
Answered by
4
F=m x a
500N=1500kg x a
a=500N/1500kg = a=0.3ms-2
Third law of motion:
2as=v2-u2
2(o.3)s=25
s=25/0.6
s=41m (appx)
Now w= f x s
= 500x 41
= 20500joules (appx)
second law of motion:
s=ut+1/2at2
41=1/2 (o.3)(t2)
0.3(t2)=82
t2=82/0.3
t2= 273
t=16.5s (appx)
now power=work done/ time taken
p=20500/16.5
power=1242Js-1 (appx)
500N=1500kg x a
a=500N/1500kg = a=0.3ms-2
Third law of motion:
2as=v2-u2
2(o.3)s=25
s=25/0.6
s=41m (appx)
Now w= f x s
= 500x 41
= 20500joules (appx)
second law of motion:
s=ut+1/2at2
41=1/2 (o.3)(t2)
0.3(t2)=82
t2=82/0.3
t2= 273
t=16.5s (appx)
now power=work done/ time taken
p=20500/16.5
power=1242Js-1 (appx)
Similar questions