/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 (I) A novice skier, starting fro... [FREE SOLUTION] | 91Ó°ÊÓ

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(I) A novice skier, starting from rest, slides down an icy frictionless 8.0\(^\circ\) incline whose vertical height is 105 m. How fast is she going when she reaches the bottom?

Short Answer

Expert verified
The skier's speed at the bottom is approximately 45.35 m/s.

Step by step solution

01

Understanding the Problem

We need to find the final velocity of a skier who slides down a frictionless incline from rest. We are given the vertical height of the hill and the angle of the incline.
02

Using Energy Conservation

Since the incline is frictionless, we use the principle of conservation of mechanical energy. The potential energy at the top will be converted into kinetic energy at the bottom.
03

Potential Energy at the Top

The potential energy at the top, relative to the bottom, is given by \( PE = mgh \), where \( m \) is the mass of the skier, \( g = 9.8 \ m/s^2 \) is the acceleration due to gravity, and \( h = 105 \ m \) is the vertical height.
04

Kinetic Energy at the Bottom

At the bottom, the potential energy is zero and the mechanical energy is all kinetic. The kinetic energy is given by \( KE = \frac{1}{2}mv^2 \), where \( v \) is the final velocity. We'll verify this energy matches the initial potential energy.
05

Setting Equations Equal

Using conservation of energy, we set the initial potential energy equal to the kinetic energy at the bottom: \( mgh = \frac{1}{2}mv^2 \). By cancelling the mass \( m \) from both sides, we have \( gh = \frac{1}{2}v^2 \).
06

Solving for Final Velocity

Rearrange the equation to solve for \( v \): \( v^2 = 2gh \). Substitute \( g = 9.8 \ m/s^2 \) and \( h = 105 \ m \), giving \( v^2 = 2 \times 9.8 \times 105 \). Calculate \( v \) by taking the square root of the result. \( v = \sqrt{2058} \approx 45.35 \ m/s \).

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Key Concepts

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

Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It allows an object to move a longer distance with less effort than lifting it vertically. In the problem of the skier, the inclined plane is the snowy slope with an angle of 8 degrees. This setup is used in the exercise to demonstrate the effects of gravity acting on the skier. Inclined planes reduce the effort needed to move an object uphill but in this exercise, we look at the motion downhill. Here, gravity helps the skier accelerate from rest. Instead of moving directly downward, the skier descends a longer, gentler path because of the incline angle. When dealing with inclined planes:
  • Gravity pulls objects directly downward, not along the slope.
  • Inclined planes experience two components of gravitational force: one acts along the slope, moving the object downhill, and the other perpendicular to the slope.
  • A key point to remember is: if the slope is frictionless, as in this problem, there is no resisting force.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. In simple terms, when an object moves, it has kinetic energy. The equation for kinetic energy is given by:\[ KE = \frac{1}{2}mv^2 \]where \( KE \) is kinetic energy, \( m \) is mass, and \( v \) is velocity. In the skier problem, as the skier moves down the frictionless incline, the potential energy at the top converts into kinetic energy. Initially, when the skier is at rest before sliding down, the kinetic energy is zero. But by the time she reaches the bottom, all the potential energy becomes kinetic energy, which makes her move faster.To determine how fast the skier is going at the bottom, we solve for velocity using the kinetic energy equation set equal to the initial potential energy, ensuring energy conservation:\[ mgh = \frac{1}{2}mv^2 \]Here, solving this equation helps us find that the kinetic energy at the bottom of the slope matches the potential energy at the top. This relationship is crucial in predicting how fast she is traveling when reaching the bottom.
Potential Energy
Potential energy is the stored energy of an object due to its position or state. For objects at a height, like the skier at the top of the hill, this is gravitational potential energy. It is calculated using the formula:\[ PE = mgh \]where \( PE \) is potential energy, \( m \) is mass, \( g \) is the acceleration due to gravity (9.8 \( m/s^2 \)), and \( h \) is height. In the context of the skier problem, the skier starts at a height of 105 meters with significant potential energy due to her elevated position. This energy is what drives her to move as she descends the hill. As she goes down, her height decreases, converting potential energy into kinetic energy.Important points about potential energy:
  • It’s directly proportional to both mass and height.
  • It helps us understand how much energy is available to convert into motion.
  • Understanding potential energy is essential to analyzing situations involving height, like sledding down a hill.

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Most popular questions from this chapter

(II) A 62-kg trampoline artist jumps upward from the top of a platform with a vertical speed of 4.5 m/s. (\(a\)) How fast is he going as he lands on the trampoline, 2.0 m below (Fig. 6-40)? (\(b\)) If the trampoline behaves like a spring of spring constant 5.8 \(\times\) 10\(^4\) N/m,how far down does he depress it?

(II) A 65-kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 23\(^\circ\) hill. The skier is pulled a distance \(x =\) 320 m along the incline and it takes 2.0 min to reach the top of the hill. If the coefficient of kinetic friction between the snow and skis is \(\mu_k =\) 0.10, what horsepower engine is required if 30 such skiers (max) are on the rope at one time?

(II) A projectile is fired at an upward angle of 38.0\(^\circ\) from the top of a 135-m-high cliff with a speed of 165 m/s. What will be its speed when it strikes the ground below? (Use conservation of energy.)

(III) Early test flights for the space shuttle used a "glider" (mass of 980 kg including pilot). After a horizontal launch at 480 km/h at a height of 3500 m, the glider eventually landed at a speed of 210 km/h. (\(a\)) What would its landing speed have been in the absence of air resistance? (\(b\))What was the average force of air resistance exerted on it if it came in at a constant glide angle of 12\(^\circ\) to the Earth's surface?

A softball having a mass of 0.25 kg is pitched horizontally at 120 km/h. By the time it reaches the plate, it may have slowed by 10%. Neglecting gravity, estimate the average force of air resistance during a pitch. The distance between the plate and the pitcher is about 15 m.

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