/*! 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 115 A \(1.50 \mathrm{~kg}\) snowball... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 115

A \(1.50 \mathrm{~kg}\) snowball is shot upward at an angle of \(34.0^{\circ}\) to the horizontal with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s}\). (a) What is its initial kinetic energy? (b) By how much does the gravitational potential energy of the snowball-Earth system change as the snowball moves from the launch point to the point of maximum height? (c) What is that maximum height?

Short Answer

Expert verified
Initial KE is 300 J, change in GPE is 93.6 J, maximum height is 6.36 m.

Step by step solution

01

Calculate Initial Kinetic Energy

The initial kinetic energy (KE) of the snowball can be calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity. Here \( m = 1.50 \mathrm{~kg} \) and \( v = 20.0 \mathrm{~m/s} \). Substitute these values into the formula to get:\[ KE = \frac{1}{2} \times 1.50 \times (20.0)^2 = 300 \mathrm{~J} \]Thus, the initial kinetic energy is 300 Joules.
02

Determine Vertical Component of Velocity

To find the change in gravitational potential energy, first determine the vertical component of the initial velocity \( v_y \). Given the initial velocity \( v = 20.0 \mathrm{~m/s} \) and the launch angle \( \theta = 34.0^\circ \), use the equation \( v_y = v \sin(\theta) \):\[ v_y = 20.0 \sin(34.0^\circ) \approx 11.17 \mathrm{~m/s} \]The vertical component of the velocity is approximately 11.17 m/s.
03

Calculate Maximum Height Using Energy Conservation

At the maximum height, the vertical component of the velocity will be zero, meaning all initial kinetic energy associated with vertical motion will convert into gravitational potential energy (GPE). The change in GPE \( \Delta U \) can also be equated to the reduction of the kinetic energy from the vertical motion.Calculate \( \Delta U \) using the formula \( \Delta U = mgh \). Since the vertical kinetic energy is given by \( \frac{1}{2}mv_y^2 \), and KE in the vertical direction equals the change in GPE, we have:\[ \frac{1}{2} \times 1.50 \times (11.17)^2 = 1.50 \times 9.81 \times h \]Solving for \( h \):\[ h = \frac{\frac{1}{2} \times 1.50 \times (11.17)^2}{1.50 \times 9.81} \approx 6.36 \mathrm{~m} \]Therefore, the maximum height is approximately 6.36 meters.
04

Change in Gravitational Potential Energy

The change in gravitational potential energy \( \Delta U \) is equal to \( mgh \), where \( h \) is the maximum height calculated:\[ \Delta U = 1.50 \times 9.81 \times 6.36 \approx 93.6 \mathrm{~J} \]Thus, the change in gravitational potential energy as the snowball reaches its maximum height is approximately 93.6 Joules.

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Key Concepts

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

Gravitational Potential Energy
Gravitational Potential Energy (GPE) is the energy that an object possesses due to its position in a gravitational field. In simpler terms, it's the energy an object has because of its height above the ground. When the snowball in our exercise is launched, it begins to gain height.
  • This builds up its gravitational potential energy.
  • The higher the snowball goes, the more GPE it gains.
  • GPE is calculated using the formula: \( U = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)), and \( h \) is the height.
In the case of our snowball, when it reaches the maximum height of approximately 6.36 meters, its GPE compared to the starting point is about 93.6 Joules. This is precisely the energy that has been transferred from kinetic energy, which brings us to the next concept.
Conservation of Energy
The principle of conservation of energy is a key concept in physics. It states that energy cannot be created or destroyed, only transformed from one form to another.
  • In the scenario with our snowball, the energy transformation is quite clear.
  • As the snowball is launched, it has a certain amount of kinetic energy (KE), which is determined by its speed and mass.
  • When kinetic energy is converted into gravitational potential energy, the total mechanical energy remains constant.
Initially, the snowball's kinetic energy was computed to be 300 Joules. As it ascends, some of this energy converts into gravitational potential energy, causing the snowball to slow down until it reaches its maximum height where all the vertical kinetic energy has been converted. Thus, both forms of energy work in harmony, conserving total energy throughout the motion.
Projectile Motion
Projectile motion involves objects that are thrown or propelled into the air and affected by gravity.
  • The path taken by such a projectile is typically curved, known as a trajectory.
  • It is usually divided into horizontal and vertical components of motion.
Our snowball was launched at an angle of 34 degrees to the horizontal with a speed of 20 m/s. The horizontal and vertical components can be separately analyzed to understand its trajectory.
  • The horizontal motion is uniform, meaning it moves at a constant speed since no horizontal forces are acting on it, except for air resistance which is often neglected in basic physics problems.
  • The vertical motion, however, is influenced by gravity, causing the snowball to move upwards until its velocity zeroes out at the peak height before descending.
Ultimately, understanding projectile motion allows us to predict an object's path and calculate critical points like maximum height and range.

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Most popular questions from this chapter

A block is sent sliding down a frictionless ramp. Its speeds at points \(A\) and \(B\) are \(2.00 \mathrm{~m} / \mathrm{s}\) and \(2.60 \mathrm{~m} / \mathrm{s}\), respectively. Next, it is again sent sliding down the ramp, but this time its speed at point \(A\) is \(4.00 \mathrm{~m} / \mathrm{s}\). What then is its speed at point \(B ?\)

The pulley has negligible mass, and both it and the inclined plane are frictionless. Block \(A\) has a mass of \(1.0\) \(\mathrm{kg}\), block \(B\) has a mass of \(2.0 \mathrm{~kg}\), and angle \(\theta\) is \(30^{\circ} .\) If the blocks are released from rest with the connecting cord taut, what is their total kinetic energy when block \(B\) has fallen \(25 \mathrm{~cm} ?\)

A sprinter who weighs \(670 \mathrm{~N}\) runs the first \(7.0 \mathrm{~m}\) of a race in \(1.6 \mathrm{~s}\), starting from rest and accelerating uniformly. What are the sprinter's (a) speed and (b) kinetic energy at the end of the \(1.6 \mathrm{~s}\) ? (c) What average power does the sprinter generate during the \(1.6 \mathrm{~s}\) interval?

A \(4.0 \mathrm{~kg}\) bundle starts up a \(30^{\circ}\) incline with \(128 \mathrm{~J}\) of kinetic energy. How far will it slide up the incline if the coefficient of kinetic friction between bundle and incline is \(0.30 ?\)

An outfielder throws a baseball with an initial speed of \(81.8\) \(\mathrm{mi} / \mathrm{h}\). Just before an infielder catches the ball at the same level, the ball's speed is \(110 \mathrm{ft} / \mathrm{s}\). In foot-pounds, by how much is the mechanical energy of the ball-Earth system reduced because of air drag? (The weight of a baseball is \(9.0\) oz.)
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