/*! 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 80 A skier starts from rest and sli... [FREE SOLUTION] | 91Ó°ÊÓ

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A skier starts from rest and slides downhill. What will be the speed of the skier if he drops by 20 meters in vertical height? Ignore any air resistance (which will, in reality, be quite a lot), and any friction between the skis and the snow.

Short Answer

Expert verified
The speed of the skier is approximately 20 m/s.

Step by step solution

01

Calculate Potential Energy

First off, find the potential energy of the skier at the starting point. While the actual mass is not given, the problem rests on the principle of energy conservation, so the mass will cancel out in our equations. The formula to calculate potential energy is \(PE = m*g*h\). Here, \(m\) is the mass of the skier, \(g\) is the acceleration due to gravity (9.8 m/s^2) and \(h\) is the height (20 m). Plugging in these values we get \(PE = m*9.8*20\).
02

Conservation of Energy

Energy is conserved in this scenario, so the potential energy at the start will equal the kinetic energy at the end, since all the potential energy will be converted into kinetic energy. The formula for kinetic energy is \(KE = 0.5*m*v^2\), where \(m\) is mass and \(v\) is velocity, the value which we need to find. So we can equate our kinetic and potential energy equations: \(m*9.8*20 = 0.5*m*v^2\).
03

Solve for v (speed)

From the equation in Step 2, we can solve for \(v\). As \(m\) appears in both sides of equation, it cancels out. The equation can be simplified as follows: \( v^2 = 2*9.8*20 \). Taking square root of both sides, we find that \( v = \sqrt{(2*9.8*20)} \).

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Key Concepts

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

Potential Energy Calculation
Understanding potential energy calculation is crucial when analyzing the energy transformations that occur in physics problems. Potential energy (PE) is the energy stored in an object due to its position or configuration. In the case of the skier in our exercise, gravitational potential energy is considered, which depends on the skier's vertical position relative to a reference level—usually the ground. The higher the skier is, the greater the potential energy.

The general formula for calculating gravitational potential energy is given by: \[ PE = m \times g \times h \]where \( m \) is the mass of the object (in this case, the skier), \( g \) is the acceleration due to gravity (standard value of 9.8 m/s² on Earth), and \( h \) is the height above the reference level.

In the provided physics problem, the potential energy for the skier can be calculated without knowing the exact mass, because it will cancel out when compared against kinetic energy later. Substituting into the potential energy formula with a height of 20 meters, the equation becomes:\[ PE = m \times 9.8 \times 20 \]This step is vital for setting the stage to apply the conservation of energy principle in the next part of the problem.
Kinetic Energy Calculation
After understanding potential energy, the focus shifts to kinetic energy (KE). Kinetic energy is the energy an object possesses because of its motion. It can be calculated using the formula:\[ KE = \frac{1}{2} \times m \times v^2 \]where \( m \) is the mass and \( v \) is the velocity of the object. This energy represents the capability of a moving object to perform work by virtue of its velocity.

In our exercise, the skier converts all the initial potential energy into kinetic energy as they slide downhill. Since we're assuming no energy is lost to friction or air resistance, conservation of energy tells us that energy is neither created nor destroyed, just converted from one form to another. Setting potential energy equal to kinetic energy and knowing the mass \( m \) will be eliminated, we equate:\[ m \times 9.8 \times 20 = \frac{1}{2} \times m \times v^2 \]Solving this equation for \( v \) is how you would determine the skier’s velocity at the bottom of the hill. The kinetic energy calculation is essential to figure out how fast something will be moving when a certain amount of potential energy has been converted to kinetic.
Basic Laws of Physics
The basic laws of physics form the foundation for solving various problems, including our skier's velocity problem. The key law applicable here is the conservation of energy principle. It states that the total energy in a closed system remains constant over time. This means that energy can change from one form to another (like potential to kinetic), but the total amount does not change.

When applied to our scenario, this law allows us to equate the skier's initial potential energy with the final kinetic energy, disregarding friction and air resistance for the sake of simplicity. The mass variable, commonly included in these energy calculations, cancels out, as seen in:\[ m \times g \times h = \frac{1}{2} \times m \times v^2 \]After ignoring mass, you're left with an equation involving just gravitational acceleration, height, and velocity, which reinforces the notion that the principles of physics apply universally, regardless of specific quantities like mass.

Understanding this and other fundamental laws of physics, such as Newton's laws of motion, is essential for students to solve problems and to grasp how the universe behaves on both large and small scales.

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Most popular questions from this chapter

A girl on a skateboard (total mass of 40 kg) is moving at a speed of \(10 \mathrm{m} / \mathrm{s}\) at the bottom of a long ramp. The ramp is inclined at \(20^{\circ}\) with respect to the horizontal. If she uavels \(14.2 \mathrm{m}\) upward along the ramp before stopping. what is the net frictional force on her?

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water with an initial speed of \(15.0 \mathrm{m} / \mathrm{s}\) strikes the water with a speed of \(24.8 \mathrm{m} / \mathrm{s}\) independent of the direction thrown. (Hint: show that \(\left.K_{i}+U_{i}=K_{f}+U_{f}\right)\)

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back \(50 \mathrm{cm}\) and holds it in position with a force of 150 N. If the mass of the arrow is \(50 \mathrm{g}\) and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?

A cat's crinkle ball toy of mass \(15 \mathrm{g}\) is thrown straight up with an initial speed of \(3 \mathrm{m} / \mathrm{s}\). Assume in this problem that air drag is negligible. (a) What is the kinetic energy of the ball as it leaves the hand? (b) How much work is done by the gravitational force during the ball's rise to its peak? (c) What is the change in the gravitational potential energy of the ball during the rise to its peak? (d) If the gravitational potential energy is taken to be zero at the point where it leaves your hand, what is the gravitational potential energy when it reaches the maximum height? (e) What if the gravitational potential energy is taken to be zero at the maximum height the ball reaches, what would the gravitational potential energy be when it leaves the hand? (f) What is the maximum height the ball reaches?

In a common physics demonstration, a bowling ball is suspended from the ceiling by a rope. The professor pulls the ball away from its equilibrium position and holds it adjacent to his nose, as shown below. He releases the ball so that it swings directly away from him. Does he get struck by the ball on its return swing? What is he trying to show in this demonstration?

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