Skip to main content
Ch 10: Interactions and Potential Energy
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 10, Problem 10a

A pendulum is made by tying a 500 g ball to a 75-cm-long string. The pendulum is pulled 30° to one side, then released. What is the ball's speed at the lowest point of its trajectory?

Verified step by step guidance
1
Step 1: Identify the type of energy transformation involved. The pendulum converts potential energy at its highest point (when it is displaced by 30 degrees) into kinetic energy at the lowest point of its trajectory.
Step 2: Write the expression for gravitational potential energy at the highest point: \( U = m g h \), where \( m \) is the mass of the ball (0.500 kg), \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the height of the ball above the lowest point.
Step 3: Determine the height \( h \) using trigonometry. The string length is \( L = 0.75 \) m, and the pendulum is displaced by 30 degrees. The vertical height \( h \) can be calculated as \( h = L - L \cos(\theta) \), where \( \theta = 30 \) degrees.
Step 4: At the lowest point, all the potential energy is converted into kinetic energy. Write the expression for kinetic energy: \( K = \frac{1}{2} m v^2 \), where \( v \) is the speed of the ball at the lowest point. Set the potential energy equal to the kinetic energy: \( m g h = \frac{1}{2} m v^2 \).
Step 5: Solve for \( v \) by canceling \( m \) from both sides and rearranging the equation: \( v = \sqrt{2 g h} \). Substitute the values of \( g \) and \( h \) (calculated in Step 3) to find the speed of the ball at the lowest point.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Conservation of Energy

The principle of conservation of energy states that in a closed system, the total energy remains constant. In the case of the pendulum, the potential energy at the highest point (when pulled to the side) is converted into kinetic energy at the lowest point. This relationship allows us to calculate the speed of the ball at the lowest point using the initial height and mass.
Recommended video:
Guided course
06:24
Conservation Of Mechanical Energy

Potential Energy

Potential energy is the energy stored in an object due to its position in a gravitational field. For the pendulum, when the ball is raised to a certain height, it gains gravitational potential energy, which can be calculated using the formula PE = mgh, where m is mass, g is the acceleration due to gravity, and h is the height above the lowest point.
Recommended video:
Guided course
07:24
Potential Energy Graphs

Kinetic Energy

Kinetic energy is the energy of an object in motion, defined by the formula KE = 1/2 mv², where m is mass and v is velocity. At the lowest point of the pendulum's swing, all the potential energy has been converted into kinetic energy, allowing us to determine the speed of the ball by equating the potential energy at the starting height to the kinetic energy at the lowest point.
Recommended video:
Guided course
06:07
Intro to Rotational Kinetic Energy
Related Practice
Textbook Question
The spring in FIGURE EX10.21a is compressed by 10 cm. It launches a block across a frictionless surface at 0.50 m/s. The two springs in Figure EX10.21b are identical to the spring of Figure EX10.21a. They are compressed by the same 10 cm and launch the same block. What is the block's speed now?
1655
views
Textbook Question

The maximum energy a bone can absorb without breaking is surprisingly small. Experimental data show that a leg bone of a healthy, 60 kg human can absorb about 200 J. From what maximum height could a 60 kg person jump and land rigidly upright on both feet without breaking his legs? Assume that all energy is absorbed by the leg bones in a rigid landing.

3448
views
Textbook Question

A 55 kg skateboarder wants to just make it to the upper edge of a 'quarter pipe,' a track that is one-quarter of a circle with a radius of 3.0 m. What speed does he need at the bottom?

2096
views
Textbook Question

A 1500 kg car traveling at 10 m/s suddenly runs out of gas while approaching the valley shown in FIGURE EX10.11. The alert driver immediately puts the car in neutral so that it will roll. What will be the car's speed as it coasts into the gas station on the other side of the valley? Ignore rolling friction.

2777
views
1
rank
Textbook Question

A 20 kg child is on a swing that hangs from 3.0-m-long chains. What is her maximum speed if she swings out to a 45° angle?

2385
views
1
rank
Textbook Question

In a hydroelectric dam, water falls 25 m and then spins a turbine to generate electricity. What is ΔUG\(\Delta\) U_{G} of 1.0 kg of water?

2224
views