/*! 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 119 A \(50 \mathrm{~g}\) ball is thr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(50 \mathrm{~g}\) ball is thrown from a window with an initial velocity of \(8.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) above the horizontal. Using energy methods, determine (a) the kinetic energy of the ball at the top of its flight and (b) its speed when it is \(3.0 \mathrm{~m}\) below the window. Does the answer to (b) depend on either (c) the mass of the ball or (d) the initial angle?

Short Answer

Expert verified
(a) KE_top = 1.20 J; (b) Speed = 9.17 m/s; (c) No; (d) No, angle does not affect speed at given height.

Step by step solution

01

Finding initial kinetic energy

The initial kinetic energy (KE) can be calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]Where \( m = 0.050 \text{ kg} \) (mass) and \( v = 8.0 \text{ m/s} \) (initial velocity). Calculating it gives:\[ KE = \frac{1}{2} \times 0.050 \times (8.0)^2 = 1.6 \text{ J} \]
02

Determine vertical component of velocity

To evaluate the kinetic energy at the top, determine the initial vertical velocity component:\[ v_{y0} = v_0 \sin(\theta) = 8.0 \sin(30^{\circ}) = 4.0 \text{ m/s} \]
03

Determine height at the top of the flight

The ball reaches the maximum height when the vertical velocity component is zero, \(v_y = 0\). Use energy conservation:\[ \text{Potential Energy Gain} = \text{Initial Vertical KE Loss} \] \[ mgh = \frac{1}{2}mv_{y0}^2 \] Simplifying gives:\[ gh = \frac{1}{2}v_{y0}^2 \] \[ h = \frac{(4.0)^2}{2 \times 9.8} = 0.818 \text{ m} \]
04

Calculate KE at top of the flight

At the top, only horizontal kinetic energy remains. The horizontal component of the velocity is constant:\[ v_{x0} = v_0 \cos(\theta) = 8.0 \cos(30^{\circ}) = 6.93 \text{ m/s} \] The kinetic energy at the top is now:\[ KE_{\text{top}} = \frac{1}{2}m(v_{x0})^2 = \frac{1}{2} \times 0.050 \times (6.93)^2 = 1.20 \text{ J} \]
05

Use energy conservation to find speed at 3.0m below

At 3.0 m below the initial point, use conservation of energy:\[ KE_{\text{initial}} + PE_{\text{initial}} = KE + PE \]Initially, KE_initial is 1.6 J and PE_initial is zero. At 3.0m below, PE is \(-mgh\):\[ 1.6 + 0 = \frac{1}{2}mv^2 - (0.050 \times 9.8 \times 3.0) \] \[ 1.6 + 0.50 \approx \frac{1}{2} \times 0.050 \times v^2 \] \[ 2.1 = 0.025v^2 \] \[ v^2 = \frac{2.1}{0.025} \] \[ v = \sqrt{84} \approx 9.17 \text{ m/s} \]
06

Influence of mass and angle

(c) The speed of the ball at 3.0m below the window does not depend on the mass because it cancels out in the energy equations due to its linearity. (d) The answer does not directly depend on the initial angle; it influences the trajectory but not the energy conservation equation at any height when only magnitude of velocity is considered.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy
Kinetic energy is a significant concept when discussing objects in motion. It refers to the energy possessed by an object due to its motion. For an object with mass, its kinetic energy (KE) can be calculated using the formula:
  • \[ KE = \frac{1}{2}mv^2 \]
Here, \( m \) represents the mass of the object and \( v \) denotes its velocity. An increase in either mass or velocity will result in higher kinetic energy. This energy is maximal just after an object starts moving and is pivotal in defining the object's capacity to perform work due to its motion.

For example, at the beginning of its flight, a thrown ball has a significant amount of kinetic energy. As it rises, the vertical component of its velocity slows down due to gravity, affecting its kinetic energy. Once the ball reaches the highest point in its trajectory, the vertical component of the velocity becomes zero. However, any horizontal motion continues to contribute to the kinetic energy at this peak, hence only the horizontal component plays a role in calculating the kinetic energy at this point.
Potential Energy
Potential energy is another crucial form of energy that is defined as the energy held by an object because of its position in a force field, commonly a gravitational field. The basic formula for gravitational potential energy (PE) is:
  • \[ PE = mgh \]
where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above some reference point.

In the given exercise, potential energy increases as the ball ascends because it gains height. This gain comes from converting the initial kinetic energy into potential energy due to the upward motion against gravity. When calculating potential energy, it's important to know that it depends on the object's height. Thus, when the ball returns to 3.0 m below the window, its potential energy becomes negative, indicating that it falls below the initial point of reference. This negative potential energy contributes to the total energy calculations, but doesn't directly define how fast the ball might be moving downward.
Velocity Components
Understanding velocity components is integral when analyzing the motion of an object in two dimensions. Velocity, which is a vector quantity, has both magnitude and direction. When an object like a ball is thrown at an angle, its velocity can be broken down into two components - horizontal and vertical.
  • The horizontal component \( v_{x} \) is given by \( v \cos(\theta) \), where \( \theta \) is the angle above the horizontal.
  • The vertical component \( v_{y} \) is calculated by \( v \sin(\theta) \).
The horizontal component remains constant if air resistance is negligible, as there is no horizontal force acting to change it. This means that it contributes continuously to the kinetic energy.

The vertical component, on the other hand, changes due to gravitational acceleration. It starts as a portion of the initial velocity, slows down to zero at the peak, and then increases in the opposite direction as the object falls back down. These components are essential in understanding the object's trajectory and calculating energy changes throughout the motion, as demonstrated in the exercise for identifying the kinetic energy at different heights.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

Each second, \(1200 \mathrm{~m}^{3}\) of water passes over a waterfall \(100 \mathrm{~m}\) high. Three-fourths of the kinetic energy gained by the water in falling is transferred to electrical energy by a hydroelectric generator. At what rate does the generator produce electrical energy? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(1000 \mathrm{~kg}\).)

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 \(0.30 ?\)

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant \(k=400 \mathrm{~N} / \mathrm{m} ;\) the other end of the spring is fixed in place. The cookie has a kinetic energy of \(20.0 \mathrm{~J}\) as it passes through the spring's equilibrium position. As the cookie slides, a frictional force of magnitude \(10.0 \mathrm{~N}\) acts on it. (a) How far will the cookie slide from the equilibrium position before coming momentarily to rest? (b) What will be the kinetic energy of the cookie as it slides back through the equilibrium position?

A \(70 \mathrm{~kg}\) firefighter slides, from rest, \(4.3 \mathrm{~m}\) down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that the frictional force of the pole on her is negligible, what is her speed just before reaching the ground floor? (b) If the firefighter grasps the pole more firmly as she slides, so that the average frictional force of the pole on her is \(500 \mathrm{~N}\) upward, what is her speed just before reaching the ground floor?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.