Question

car weighing 1000 kg climbs up a hill of slope 1 in 98 with a velocity of 72 KMPH. If the power delivered by the engine is 3.5 kw, the frictional force acting on the car is 25 N 175 N 75 N 125 N

Answer 1

Explanation:

slope 1 in 98 clearly means that sin x is 1 / 98

mg . sin x and f ( force of friction ) is balanced by opposite force which is F

so according to this statement we can say that,

F = mg . sin x + f

F = 1000 × 9.8 × 1 / 98 + f

F = ( 100 + f ) N

Power = 3.5kW

velocity = 72 km /hr or we can say 20 m/ s

P = W / t

but W = F . s

P = F.s / t

and s / t = v

so we have power in terms of velocity

P = F × v

P = ( 100 + f ) × 20

3.5 × 10³ = 20 ( 100 + f )

3.5 × 10² = 2 ( 100 + f )

350 = 200 + 2f

2f = 150

f = 75 N

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