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Mariulka [41]
3 years ago
7

When an automobile moves with constant velocity, the power developed is used to overcome the frictional forces exerted by the ai

r and the road. If the engine develops 30 hp, what total frictional force acts on the car at 140 mph? One horsepower equals 746 W, and one mile is 1609 m. Answer in units of N.
Physics
2 answers:
cluponka [151]3 years ago
8 0

Answer:

Explanation:

Given:

Power, p = 30 hp

1 hp = 746 W

= 22.38 kW

Velocity, v = 140 mph

1 mile = 1609 m

Converting mph to m/s,

140 mile/hour × 1609 m/1 mile × 1 hour/3600 s

= 62.572 m/s

Summation of forces = 0

Power, p = energy/time

Energy = force × distance;

Velocity = distance/time;

Power = force × velocity

Force, fr = 22380/62.572

= 357.67 N

Frictional force, fr = 0.358 kN.

geniusboy [140]3 years ago
6 0

Answer:

The total frictional force is 358.0 newtons

Explanation:

Power is the amount of average work (W) an object does on a period of time (Δt):

P=\frac{W}{\Delta t}

Remember average work is average force (F) times displacement (Δs):

P=\frac{F \Delta s}{\Delta t}

but displacement over time is average speed v=\frac{\Delta s}{\Delta t}, then:

P=Fv (1)

That is, the power of the car is the force the engine does times the speed of the car. As the question states, if the car is at constant velocity then the power developed is used to overcome the frictional forces exerted by the air and the road, that is by Newton's first law, the force the motor of the car does is equal the force of frictional forces. So, to find the frictional forces we only have to solve (1) for F:

F=\frac{P}{v}

Knowing that 1hp is 746W then 30hp=22380W and 1 mile = 1609m then 140 mph = 225308\frac{m}{h} = 62.5 \frac{m}{s}, then:

F=\frac{22380}{62.5}=358.0 N

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3 years ago
A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
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Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

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So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

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when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

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we can divide both sides of the equation into mg so we get:

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\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

which yields:

α=19.29°

which rounds to:

α=19°

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3 years ago
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