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Romashka [77]
3 years ago
15

A 65-kg skier grips a moving rope that is powered by an engine and is pulled at a constant speed to the top of a 230 hill. The s

kier is pulled a distance x = 320 m along the incline and it takes 2.0 minutes to reach the top of the hill. If the coefficient of kinetic friction between the snow and skis is µk = 0.10, what horsepower engine is required if 30 such skiers are on the rope at one time? 1 hp = 746 W
Physics
1 answer:
Sveta_85 [38]3 years ago
3 0

Answer:

The required power by the engine is 33.0 hp

Explanation:

Solution

Newton's second law says that, the net force Fnet on an object of mass m will accelerates the object

Where

Fnet = ma

a = acceleration

θ = angle of incline,

m = mass of the 30 skiers,

f = frictional force

N = normal force

mg sinθ, mg cos θ are components of weight skier

F = the force applied by engine

Now,

The skier mass  is 65 kg

We calculate the mass of the 30 skier

m = 30 (65kg) = 1950 kg

Calculate the net force acting on the skiers along the x-axis

Fnet, x=ma

Now,

F-mg sin θ - f = 0

F= mg sin θ + f -----(1)

The kinetic frictional force is denoted by

f = μk N ------(2)

μk = The coefficient of the kinetic friction

We now, calculate the net force acting on the skiers along y axis

Fnet, y = ma

N- mg cos θ = 0

so,

N = mg cos θ

This value is  substituted in equation (2)

f = μk mg cos θ

we substitute the value for equation (1)

F = mg sin θ + μk mg cos θ

mg =  sinθ + μk cos θ)-----(3)

The next step is to calculate the work done by the engine in pulling the skiers, the incline top by applying the equation 3

W = Fx

= mg ( sinθ + μk cos θ)x

x = the displacement

we now substitute 1950 kg for m, 23° for θ, 0.10 for μk and 320m for x

so,

W = mg ( sinθ + μk cos θ)x

= (1950 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (320 m)

= 2.99 * 10 ^6 J

Then,

The time from minute to s is converted

t =(2.0min) ( 60sec/1.0min) = 120 sec

Now we calculate the power needed by the engine to pull the skiers at the incline top

Thus,

P = W/t

we substitute  2.955 * 10 ^6 J for W and  120 s for t

we have,

P = 2.955 * 10 ^ 6 J/ 120 s

= ( 2.4625 * 10 ^ 4 W) (1.0 hp/746 W)

= 33.0 hp

In conclusion the required power by the engine is 33.0 hp

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distance of ship from center of the ring, r =31.0 m

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Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N

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