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Standard VIII

Physics

Mechanical Advantage

When an autom...

Question

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the car by the air and the road. If the power developed by an engine is $175 h p$ , estimate the total friction force (approx.) acting on the car when it is moving at a speed of $25 m / s$ . One horsepower equals $746 W$ .

2.50 k N

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3.60 k N

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3.73 k N

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5.00 k N

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Solution

The correct option is D $5.00 k N$
Given, power developed by the engine,
$P = 175 h p = (175 \times 746) W$
Speed of the car, $v = 25 m / s$
Total friction force acting on the car, $f = \frac{P}{v}$
$f = \frac{175 \times 746}{25} = 5222 N = 5.22 k N$
$f \approx 5.00 k N$
Final answer: (d)

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An engine pull a 1500kg of car on a level road at a constant speed of 5m/s against a frictional force of 500N. Calculate the power expended by the engine. What extra power has the engine to expend in order to maintain the same speed of the car up an inclined plane having a gradient of 1in 10?

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Mechanical Advantage

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The correct option is D 5.00 kNGiven, power developed by the engine, P=175 hp=(175×746)W Speed of the car, v=25 m/s Total friction force acting on the car, f=Pv f=175×74625=5222 N=5.22 kN f≈5.00 kN Final answer: (d)

The correct option is D $5.00 k N$
Given, power developed by the engine,
$P = 175 h p = (175 \times 746) W$
Speed of the car, $v = 25 m / s$
Total friction force acting on the car, $f = \frac{P}{v}$
$f = \frac{175 \times 746}{25} = 5222 N = 5.22 k N$
$f \approx 5.00 k N$
Final answer: (d)