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A ski tow operates on a 15.0\(^\circ\) slope of length 300 m. The rope moves at 12.0 km/h and provides power for 50 riders at one time, with an average mass per rider of 70.0 kg. Estimate the power required to operate the tow.

Short Answer

Expert verified
The power required is approximately 29.66 kW.

Step by step solution

01

Convert the Speed of the Rope

First, we need to convert the speed of the rope from km/h to m/s. Since 1 km = 1000 meters and 1 hour = 3600 seconds, the conversion formula is:\[\text{Speed in m/s} = \text{Speed in km/h} \times \frac{1000}{3600} \approx 3.33 \text{ m/s}.\]
02

Determine the Total Mass

Next, calculate the total mass of all riders. Given that there are 50 riders, each with an average mass of 70.0 kg, the total mass is:\[\text{Total mass} = 50 \times 70.0 \text{ kg} = 3500 \text{ kg}.\]
03

Calculate the Gravitational Force Component

To find the component of gravitational force acting along the slope, we use:\[F = m \cdot g \cdot \sin(\theta),\]where:- \( m = 3500 \text{ kg} \) (total mass),- \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity),- \( \theta = 15.0^\circ \) (slope angle).First, calculate \( \sin(15.0^\circ) \):\[\sin(15.0^\circ) \approx 0.2588.\]So the force is:\[F = 3500 \times 9.81 \times 0.2588 \approx 8906 \text{ N}.\]
04

Calculate the Power Required

Power is calculated by multiplying the force by the speed at which it is applied:\[P = F \times v,\]where:- \( F = 8906 \text{ N} \) (force calculated in the previous step),- \( v = 3.33 \text{ m/s} \) (speed of the rope).Substitute the values:\[P = 8906 \times 3.33 \approx 29660 \text{ W}.\]
05

Convert Power to Kilowatts

Finally, convert the power from watts to kilowatts. Since 1 kilowatt = 1000 watts:\[P = \frac{29660}{1000} \approx 29.66 \text{ kW}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

gravitational force calculation
When dealing with gravitational force calculation, it’s important to understand the influence of gravity on different bodies. In this particular problem, the gravitational force acts along the slope where the ski tow moves. This scenario helps us see gravity's impact in combination with angles.
We calculate the force component acting parallel to the slope using the formula:
  • \( F = m \cdot g \cdot \sin(\theta) \)
where:
  • \( m \) is the total mass of all riders,
  • \( g \) is the acceleration due to gravity, approximately 9.81 m/s\(^2\),
  • \( \theta \) is the angle of the slope.
The sine function gives us the ratio of the opposite side to the hypotenuse in a right triangle, aligning with the slope's inclination. This calculation isolates how much gravitational force pulls the riders down the slope.
To understand and perform this calculation right, students should visualize triangles formed by slope setups and recognize how gravity splits into components: one along the slope and one perpendicular.
power calculation
Power calculation in physics often relates to how energy is employed over time. In the ski tow problem, we determine how much power is necessary to move riders against the gravitational pull along the slope. Power quantifies the rate at which work is done. Here, work is overcoming the gravitational force to move uphill.
We calculate power with the equation:
  • \( P = F \times v \)
where:
  • \( F \) is the gravitational force applied.
  • \( v \) is the velocity of the rope.
By substituting the correct values into the equation, we find how much power is needed by the ski tow to move uphill. The conversion from watts to kilowatts is a simple division by 1000, useful for scaling our answer meaningfully.
Clear understanding of power helps in various applications beyond ski tows, such as determining engine sizes for vehicles or energy consumption of appliances.
kinematics and dynamics
Kinematics and dynamics are branches of mechanics that describe the motion of objects. In this exercise, dynamics play a crucial role as we're interested in the forces acting on the ski tow and how this affects movement.
The rope moves constantly at 12.0 km/h, which we convert to meters per second using:
  • \( \frac{1000}{3600} \) factor to change km/h to m/s.
This speed is not just a number; it represents how quickly energy needs to counter gravitational forces to move the skiers uphill.
Combining kinematics with dynamics helps in understanding:
  • The influence of speed and mass on system energy requirements,
  • The equilibrium of forces ensuring balanced movement, neither accelerating nor slowing.
Mastering these concepts builds the foundation for more complex analyses in fields like engineering and aeronautics.
energy conversion
Energy conversion is a concept pivotal in understanding how machines power different systems. In our ski tow problem, we convert electrical or mechanical energy into kinetic energy to move riders uphill.
This conversion involves:
  • Using power derived from a motor or engine to apply force along the slope,
  • Overcoming gravitational pull to keep the rope moving with constant velocity.
Understanding energy conversion in this context underscores the practical application of physics principles. It allows systems like ski tows, or even roller coasters, to function efficiently and safely.
Developing a grasp on how energy changes forms and the efficiencies involved is crucial in optimizing various technological processes.
trigonometry in physics
Trigonometry in physics simplifies complex relationships involving angles, distances, and forces. In the ski tow scenario, trigonometry helps determine gravitational force components.
The angle \( \theta \) (15.0° in this problem) is essential in calculating how much force acts along and perpendicular to the slope. Using trigonometric functions like sine:
  • \( \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \)
explains how to project gravitational force onto different axes paralleling the slope.
When solving physics problems, recognizing when to apply trigonometry can demystify complex situations. Trigonometry serves in breaking down forces, analyzing waves, and predicting projectile paths, making it a versatile tool in physics problem-solving.

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Most popular questions from this chapter

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?

Use the work\(-\)energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0-m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 m into the air. How fast was the boulder moving just as it left the volcano?

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