Ch 09: Work and Kinetic Energy

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 09: Work and Kinetic Energy

Problem 44b

Chapter 9, Problem 44b

Doug uses a 25 N horizontal force to push a 5.0 kg crate up a 2.0-m-high, 20° frictionless slope. What is the speed of the crate at the top of the slope?

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Identify the forces acting on the crate: The applied force (25 N), the gravitational force, and the normal force. Since the slope is frictionless, there is no frictional force.

Break the gravitational force into components parallel and perpendicular to the slope. The parallel component is given by \( F_{\text{gravity, parallel}} = m g \sin(\theta) \), where \( m = 5.0 \ \text{kg} \), \( g = 9.8 \ \text{m/s}^2 \), and \( \theta = 20^\circ \).

Determine the net force acting along the slope: \( F_{\text{net}} = F_{\text{applied}} - F_{\text{gravity, parallel}} \). Use the values of \( F_{\text{applied}} = 25 \ \text{N} \) and the calculated \( F_{\text{gravity, parallel}} \).

Use Newton's second law to find the acceleration of the crate along the slope: \( a = \frac{F_{\text{net}}}{m} \). Substitute the net force and the mass of the crate to calculate \( a \).

Apply the kinematic equation \( v^2 = u^2 + 2 a d \) to find the speed \( v \) of the crate at the top of the slope. Here, \( u = 0 \ \text{m/s} \) (initial speed), \( a \) is the acceleration calculated in the previous step, and \( d \) is the distance along the slope, which can be found using trigonometry: \( d = \frac{h}{\sin(\theta)} \), where \( h = 2.0 \ \text{m} \).

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Key Concepts

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

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This principle is crucial for analyzing the motion of the crate as it is pushed up the slope, allowing us to calculate the net force and resulting acceleration.

Work-Energy Principle

The Work-Energy Principle asserts that the work done on an object is equal to the change in its kinetic energy. In this scenario, the work done by Doug's force will contribute to the crate's kinetic energy as it moves up the slope, enabling us to determine its speed at the top.

Kinematics

Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. Understanding kinematic equations is essential for calculating the final speed of the crate after it has been pushed up the slope, given its initial conditions and the work done on it.

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Textbook Question

A 150 g particle at x = 0 is moving at 2.00 m/s in the + x - direction. As it moves, it experiences a force given by Fₓ = (0.250 N) sin (x/2.00 m). What is the particle's speed when it reaches x = 3.14 m?

Textbook Question

A 1000 kg elevator accelerates upward at 1.0 m/s² for 10 m, starting from rest. How much work does the tension in the elevator cable do on the elevator?

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Susan's 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30° above the floor. The tension is a constant 30 N and the coefficient of friction is 0.20. Use work and energy to find Paul's speed after being pulled 3.0 m.

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A pile driver lifts a 250 kg weight and then lets it fall onto the end of a steel pipe that needs to be driven into the ground. A fall from an initial height of 1.5 m drives the pipe in 35 cm. What is the average force that the weight exerts on the pipe?

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Textbook Question

The energy used to pump liquids and gases through pipes is a significant fraction of the total energy consumption in the United States. Consider a small volume V of a liquid that has density ρ. Assume that the fluid is nonviscous so that friction with the pipe walls can be neglected. An upward-pushing force from a pump lifts this volume of fluid a height h at constant speed. How much work does the pump do?

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A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from rest to 20 m/s. What is the average power output of each? Average power over a time interval ∆t is ∆E/∆t.

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