Ask question

Ask question

All categories

2 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

1 answer:

Sveta_85 [38]2 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

You might be interested in

Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padd

Free_Kalibri [48]

Answer:

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

We'll take absolute value as;

v_f = 0.87 m/s

5 0

3 years ago

A force of 45 N is applied tangentially to the rim of a solid disk of radius 0.12 m. The disk rotates about an axis through its

bearhunter [10]

Answer:

Mass of the disk will be 2.976 kg

Explanation:

We have given force F = 45 N

Radius of the disk r = 0.12 m

Angular acceleration $\alpha =140rad/sec^2$

We know that torque $\tau =I\alpha$

And $\tau =Fr$

So $Fr=I\alpha$ , here I is moment of inertia

So $50\times 0.12=I\times 140$

$I=0.0428kgm^2$

We know that moment of inertia $I=\frac{1}{2}mr^2$

So $0.0428=\frac{1}{2}\times m\times 0.12^2$

m = 2.976 kg

6 0

3 years ago

John argues that energy is not conserved because when someone presses the brakes in a moving car, kinetic energy is lost. What s

babunello [35]

kinetic energy is converted into elastic potential energy stored in the brakes.

7 0

2 years ago

Consider the plot below describing motion along a straight line with an initial position of 10 m. −4 −3 −2 −1 0 1 2 3 4 5 6 1 2

Ugo [173]

Answer:

$+1 m/s^2$

Explanation:

Acceleration is given by

$a=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in velocity

$\Delta t$ is the time interval in which the change in velocity occurs

To find the acceleration at 1 second, we can take the data at t = 1 s and t = 2. We find:

$\Delta t = -3 -(-4) = 1 s$

$\Delta v = 2 - (1) = +1 m/s$

So, the acceleration is

$a=\frac{+1}{1}=+1 m/s^2$

4 0

3 years ago

What distance is covered by an airplane traveling at a velocity of 660 miles per hour in 3.5 hours?

N76 [4]

As per the question, the velocity of the airplane [v] = 660 miles per hour.

The total time taken by airplane [t] = 3.5 hours.

We are asked to determine the total distance travelled by the airplane during that period.

The distance covered [ S] by a body is the product of velocity with the time.

Mathematically distance covered = velocity × total time

S = v × t

= 660 miles/hour ×3.5 hours

= 2310 miles.

Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.

4 0

2 years ago