/*! 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 69 A high jumper approaches the bar... [FREE SOLUTION] | 91Ó°ÊÓ

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A high jumper approaches the bar at \(9.0 \mathrm{~m} / \mathrm{s}\). What is the highest altitude the jumper can reach, if he does not use any additional push off the ground and is moving at \(7.0 \mathrm{~m} / \mathrm{s}\) as he goes over the bar?

Short Answer

Expert verified
Based on the conservation of energy principle, the highest altitude the high jumper can reach is approximately 1.02 meters.

Step by step solution

01

Write down the energy conservation equation

In this step, we need to write the energy conservation equation \(K_f + U_f = K_i + U_i\) as \(K_i - K_f = U_f - U_i\).
02

Calculate the initial and final kinetic energy

Using the given initial velocity (\(v_i = 9.0 \mathrm{~m} / \mathrm{s}\)) and final velocity (\(v_f = 7.0 \mathrm{~m} / \mathrm{s}\)), we will now calculate the initial and final kinetic energy using the formula \(K=\frac{1}{2}mv^2\). Due to conservation of mass, we can divide by mass \(m\) in the calculations, which cancels it out. \(K_i = \frac{1}{2}m v_i^2 = \frac{1}{2}m(9.0)^2\) \(K_f = \frac{1}{2}m v_f^2 = \frac{1}{2}m(7.0)^2\)
03

Substitute the calculated values in the energy conservation equation

Now, we substitute the values of \(K_i\) and \(K_f\) calculated in Step 2 and the initial potential energy into the energy conservation equation. Since \(U_i = 0\), we only need to consider the final potential energy, \(U_f = mgh\): \(\frac{1}{2}m(9.0)^2 - \frac{1}{2}m(7.0)^2 = mgh\)
04

Solve for the highest altitude (h)

In this step, we will solve for \(h\). We can first cancel out the mass \(m\) from both sides of the equation: \((9.0)^2 - (7.0)^2 = 2gh\) Now, we can solve for \(h\): \(h = \frac{(9.0)^2 - (7.0)^2}{2g}\) Using the value of acceleration due to gravity, \(g = 9.81 \mathrm{~m} / \mathrm{s^2}\), we can calculate the highest altitude: \(h = \frac{(9.0)^2 - (7.0)^2}{2(9.81)} \approx 1.02 \mathrm{~m}\) The highest altitude the jumper can reach is approximately \(1.02 \mathrm{~m}\).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It's an important concept in physics because it quantifies the amount of work that a moving object can perform due to its velocity. The formula for kinetic energy (\textbf{KE}) is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.

In the context of our high jumper problem, we can see kinetic energy in action as the jumper moves towards the bar. As the velocity of the jumper changes, so does his kinetic energy. This is crucial, because, upon taking off from the ground, some of the kinetic energy is converted to potential energy, allowing the jumper to reach a certain height.
Potential Energy
Potential energy is the energy stored in an object due to its position or arrangement. Gravitational potential energy, for instance, is energy that an object possesses because of its position in a gravitational field. The formula for gravitational potential energy (\textbf{PE}) is given by \( PE = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (\( 9.81 \text{m/s}^2 \) on Earth), and \( h \) is the height above the reference point.

For our high jumper, the highest altitude he can reach is determined by when all the kinetic energy, that isn’t lost to other forces, is converted to potential energy. The jumper's initial potential energy at ground level is zero which increases as he ascends.
Conservation of Mechanical Energy
The principle of conservation of mechanical energy states that the total mechanical energy of an isolated system remains constant as long as only conservative forces, like gravity, are acting on it. Mechanical energy is the sum of kinetic and potential energy. This principle can be written as \( KE_{i} + PE_{i} = KE_{f} + PE_{f} \), with ‘i’ indicating initial and ‘f’ indicating final states. Non-conservative forces, like friction or air resistance, can cause this energy to change, but they are ignored in ideal scenarios.

In the given exercise, by setting the initial potential energy on the ground to zero and considering no other forces but gravity, we applied energy conservation to solve for the maximum height the jumper could reach. The equation \( K_f + U_f = K_i + U_i \) captures the energy transformation from kinetic to potential, allowing us to predict how high the jumper will go.
Kinematics in Physics
Kinematics is the branch of mechanics that describes the motion of points, objects, and systems without considering the forces that cause them to move. It focuses on displacement, velocity, acceleration, and time. These concepts are fundamental when analyzing the motion of objects, as they help understand trajectories and motion patterns without delving into what actually propels or alters that motion.

Relating to our jumper, kinematics would not only look at the jumper's speeds at various points but could extend to the trajectory he takes throughout his jump. Even though our problem didn’t require a deep dive into kinematics, understanding how motion is described in physics provides a foundation for more complex problems where kinematics and energy conservation intertwine.

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Most popular questions from this chapter

A mass of \(1.00 \mathrm{~kg}\) attached to a spring with a spring constant of \(100 .\) N/m oscillates horizontally on a smooth frictionless table with an amplitude of \(0.500 \mathrm{~m} .\) When the mass is \(0.250 \mathrm{~m}\) away from equilibrium, determine: a) its total mechanical energy; b) the system's potential energy and the mass's kinetic energy; c) the mass's kinetic energy when it is at the equilibrium point. d) Suppose there was friction between the mass and the table so that the amplitude was cut in half after some time. By what factor has the mass's maximum kinetic energy changed? e) By what factor has the maximum potential energy changed?

A 1.00 -kg block is resting against a light, compressed spring at the bottom of a rough plane inclined at an angle of \(30.0^{\circ}\); the coefficient of kinetic friction between block and plane is \(\mu_{\mathrm{k}}=0.100 .\) Suppose the spring is compressed \(10.0 \mathrm{~cm}\) from its equilibrium length. The spring is then released, and the block separates from the spring and slides up the incline a distance of only \(2.00 \mathrm{~cm}\) beyond the spring's normal length before it stops. Determine a) the change in total mechanical energy of the system and b) the spring constant \(k\).

Calculate the force \(F(y)\) associated with each of the following potential energies: a) \(U=a y^{3}-b y^{2}\) b) \(U=U_{0} \sin (c y)\)

A block of mass \(0.773 \mathrm{~kg}\) on a spring with spring constant \(239.5 \mathrm{~N} / \mathrm{m}\) oscillates vertically with amplitude \(0.551 \mathrm{~m}\). What is the speed of this block at a distance of \(0.331 \mathrm{~m}\) from the equilibrium position?

A variable force acting on a 0.100 - \(\mathrm{kg}\) particle moving in the \(x y\) -plane is given by \(F(x, y)=\left(x^{2} \hat{x}+y^{2} \hat{y}\right) \mathrm{N},\) where \(x\) and \(y\) are in meters. Suppose that due to this force, the particle moves from the origin, \(O\), to point \(S\), with coordinates \((10.0 \mathrm{~m},\) \(10.0 \mathrm{~m}\) ). The coordinates of points \(P\) and \(Q\) are \((0 \mathrm{~m}, 10.0 \mathrm{~m})\) and \((10.0 \mathrm{~m}, 0 \mathrm{~m})\) respectively. Determine the work performed by the force as the particle moves along each of the following paths: a) OPS b) OQS c) OS d) \(O P S Q O\) e) \(O Q S P O\)

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