/*! 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 8 A \(1.50 \mathrm{~kg}\) snowball... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(1.50 \mathrm{~kg}\) snowball is fired from a cliff \(12.5 \mathrm{~m}\) high. The snowball's initial velocity is \(14.0 \mathrm{~m} / \mathrm{s},\) directed \(41.0^{\circ}\) above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?

Short Answer

Expert verified
(a) 184.1 J; (b) -184.1 J; (c) -184.1 J.

Step by step solution

01

Identify Given Quantities

We have a snowball with mass \( m = 1.50 \, \text{kg} \), launched from a cliff of height \( h = 12.5 \, \text{m} \). Its initial velocity is \( v_0 = 14.0 \, \text{m/s} \) at an angle \( \theta = 41.0^{\circ} \) above the horizontal.
02

Compute Work Done by Gravitational Force

The work done by the gravitational force is given by \( W = mgh \), where \( g = 9.81 \, \text{m/s}^2 \). Substitute the values to find: \[ W = 1.50 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 12.5 \, \text{m} \]. Calculate this to get \( W = 184.125 \, \text{J} \).
03

Determine Change in Gravitational Potential Energy

The change in gravitational potential energy \( \Delta U \) is equal to the negative work done by gravity because \( \Delta U = -W \). Therefore, \( \Delta U = -184.125 \, \text{J} \).
04

Calculate Gravitational Potential Energy at the Ground

Since gravitational potential energy is taken to be zero at the height of the cliff, when the snowball reaches the ground, its potential energy is \(-mgh\). Calculate this using \( m = 1.50 \, \text{kg} \), \( g = 9.81 \, \text{m/s^2} \), and \( h = 12.5 \, \text{m} \). Thus, \( U = -184.125 \, \text{J} \).

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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 a concept in physics that refers to the energy an object possesses due to its position in a gravitational field. This energy is dependent on three key factors:
  • The object's mass (\(m\)
  • The gravitational acceleration (\(g\)), usually approximated as \(9.81 \, \text{m/s}^2\) on Earth's surface
  • The height (\(h\)) above a reference point
The formula for gravitational potential energy is given by \(U = mgh\).
In the context of the snowball problem, the GPE changes as the snowball moves from the cliff to the ground. When the snowball is at the top of the cliff, it has positive potential energy relative to the ground. As it falls, this energy is converted into kinetic energy, eventually reaching zero at the base if we take the ground as our reference point. It's crucial to note that potential energy is relative and can be set to zero at any convenient point, which influences how we compute changes in energy.
Mechanics
Mechanics is the branch of physics that deals with the motion of objects and the forces acting upon them. It encompasses several sub-fields, including dynamics and statics. In our exercise, we are primarily concerned with dynamics, which examines how forces, like gravity, influence motion.The work done by a force on an object can also be calculated in this realm. The formula for work done by gravity, which is the force pulling the snowball to the ground, is expressed as \(W = mgh\). This is calculated by multiplying the mass of the object, the gravitational acceleration, and the height from which the object falls.
In this scenario, the snowball's descent involves both gravitational force and its initial velocity, which comprise the core principles of mechanics. As the snowball moves, it is a perfect example of translating potential energy into kinetic energy, governed by the principles of mechanics.
Kinematics
Kinematics is a fundamental aspect of mechanics that describes the motion of objects without considering the forces that cause this motion. It's about studying the object's path, velocity, and acceleration.For our snowball example, we initially know its velocity and launch angle. The kinematics of the snowball involves understanding its trajectory as it leaves the cliff:
  • The initial velocity (\(v_0 = 14.0 \, \text{m/s}\)) determines how fast it is moving.
  • The angle (\(41.0^{\circ}\)) affects how high and far it will travel.
  • Various components like horizontal and vertical displacement can be analyzed to describe the snowball's path.
By breaking down the initial velocity into vertical and horizontal components, you can predict where and when the snowball will hit the ground. This is a clear application of kinematic equations, where the interplay of velocity, time, and acceleration comes into focus.

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Most popular questions from this chapter

A \(15 \mathrm{~kg}\) block is accelerated at \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) along a horizontal frictionless surface, with the speed increasing from \(10 \mathrm{~m} / \mathrm{s}\) to \(30 \mathrm{~m} / \mathrm{s} .\) What are \((\mathrm{a})\) the change in the block's mechanical energy and (b) the average rate at which energy is transferred to the block? What is the instantaneous rate of that transfer when the block's speed is (c) \(10 \mathrm{~m} / \mathrm{s}\) and (d) \(30 \mathrm{~m} / \mathrm{s} ?\)

A worker pushed a \(27 \mathrm{~kg}\) block \(9.2 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was 0.20 , what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

A block of mass \(m=2.0 \mathrm{~kg}\) is dropped from height \(h=40 \mathrm{~cm}\) onto a spring of spring constant \(k=1960 \mathrm{~N} / \mathrm{m}\) (Fig. 8 -39). Find the maximum distance the spring is compressed.

The spring in the muzzle of a child's spring gun has a spring constant of \(700 \mathrm{~N} / \mathrm{m}\). To shoot a ball from the gun, first the spring is compressed and then the ball is placed on it. The gun's trigger then releases the spring, which pushes the ball through the muzzle. The ball leaves the spring just as it leaves the outer end of the muzzle. When the gun is inclined upward by \(30^{\circ}\) to the horizontal, a \(57 \mathrm{~g}\) ball is shot to a maximum height of \(1.83 \mathrm{~m}\) above the gun's muzzle. Assume air drag on the ball is negligible. (a) At what speed does the spring launch the ball? (b) Assuming that friction on the ball within the gun can be neglected, find the spring's initial compression distance.

131 Fasten one end of a vertical spring to a ceiling, attach a cabbage to the other end, and then slowly lower the cabbage until the upward force on it from the spring balances the gravitational force on it. Show that the loss of gravitational potential energy of the cabbage-Earth system equals twice the gain in the spring's potential energy.

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