Work And Energy Sample Problems

Work and Energy: Sample Problems and Solutions

Understanding work and energy is fundamental to grasping many concepts in physics and engineering. This comprehensive guide delves into various sample problems related to work and energy, providing detailed solutions and explanations to solidify your understanding. We'll cover different scenarios, including those involving friction, potential energy, kinetic energy, and the work-energy theorem. By the end, you'll be equipped to tackle a wide range of work and energy problems with confidence.

I. Introduction: Defining Work and Energy

Before diving into the sample problems, let's establish a clear understanding of the core concepts.

• Work (W): Work is done when a force causes an object to move in the direction of the force. It's calculated as the product of the force (F) and the displacement (d) in the direction of the force: W = Fd cos θ, where θ is the angle between the force and the displacement vectors. The SI unit for work is the Joule (J), equivalent to a Newton-meter (Nm).

• Energy (E): Energy is the capacity to do work. It exists in various forms, including:

  • Kinetic Energy (KE): The energy of motion, calculated as KE = 1/2mv², where 'm' is the mass and 'v' is the velocity.
  • Potential Energy (PE): Stored energy due to an object's position or configuration. This includes:
    • Gravitational Potential Energy (GPE): GPE = mgh, where 'g' is the acceleration due to gravity (approximately 9.8 m/s²) and 'h' is the height above a reference point.
    • Elastic Potential Energy: Energy stored in a stretched or compressed spring or other elastic material.

• Work-Energy Theorem: This theorem states that the net work done on an object is equal to the change in its kinetic energy: Wnet = ΔKE = KEf - KEi, where KEf is the final kinetic energy and KEi is the initial kinetic energy.

II. Sample Problems and Solutions

Let's tackle a series of problems demonstrating various applications of work and energy principles.

Problem 1: Calculating Work Done by a Constant Force

A worker pushes a crate weighing 50 kg across a frictionless floor with a constant horizontal force of 100 N for a distance of 5 meters. Calculate the work done by the worker.

Solution:

Here, the force (F) is 100 N, the displacement (d) is 5 m, and the angle (θ) between the force and displacement is 0° (since the force is horizontal and the displacement is horizontal).

Using the formula W = Fd cos θ:

W = (100 N)(5 m) cos 0° = 500 J

Therefore, the worker does 500 Joules of work.

Problem 2: Work Done Against Gravity

A 10 kg object is lifted vertically to a height of 2 meters. Calculate the work done against gravity.

Solution:

The force acting against gravity is the object's weight (mg), where m is the mass and g is the acceleration due to gravity (9.8 m/s²).

Force (F) = mg = (10 kg)(9.8 m/s²) = 98 N

Displacement (d) = 2 m

W = Fd cos θ = (98 N)(2 m) cos 0° = 196 J

The work done against gravity is 196 Joules.

Problem 3: Kinetic Energy and Work-Energy Theorem

A 2 kg ball is initially at rest. A constant force of 10 N acts on the ball over a distance of 5 meters. Using the work-energy theorem, calculate the final velocity of the ball.

Solution:

First, calculate the work done by the force:

W = Fd = (10 N)(5 m) = 50 J

According to the work-energy theorem, this work is equal to the change in kinetic energy:

W = ΔKE = KEf - KEi

Since the ball starts at rest, KEi = 0. Therefore:

50 J = KEf = 1/2mv²

Solving for v:

v² = (2 * 50 J) / 2 kg = 50 m²/s²

v = √50 m²/s² ≈ 7.07 m/s

The final velocity of the ball is approximately 7.07 m/s.

Problem 4: Work Done with an Angle

A child pulls a sled with a rope at an angle of 30° above the horizontal. The child applies a force of 50 N, and the sled moves 10 meters horizontally. Calculate the work done by the child.

Solution:

Here, we need to consider the component of the force that is parallel to the displacement. This component is given by Fcosθ.

Force parallel to displacement (F_parallel) = F cos θ = 50 N * cos 30° ≈ 43.3 N

Displacement (d) = 10 m

W = F_parallel * d = (43.3 N)(10 m) ≈ 433 J

The work done by the child is approximately 433 Joules.

Problem 5: Potential and Kinetic Energy Conversion

A 0.5 kg ball is dropped from a height of 10 meters. Ignoring air resistance, calculate its velocity just before it hits the ground.

Solution:

We can use the conservation of mechanical energy principle, where the total mechanical energy (potential energy + kinetic energy) remains constant. At the top, the ball has only potential energy (PE). Just before hitting the ground, it has only kinetic energy (KE).

Initial PE = mgh = (0.5 kg)(9.8 m/s²)(10 m) = 49 J

Final KE = 1/2mv² = 49 J (since energy is conserved)

Solving for v:

v² = (2 * 49 J) / 0.5 kg = 196 m²/s²

v = √196 m²/s² = 14 m/s

The velocity of the ball just before hitting the ground is 14 m/s.

Problem 6: Work Done Against Friction

A 10 kg box is pushed across a rough surface with a constant horizontal force of 20 N. The coefficient of kinetic friction between the box and the surface is 0.2. The box moves 4 meters. Calculate the work done by the applied force and the work done by friction.

Solution:

First, calculate the force of friction:

Friction force (Ff) = μk * N = μk * mg = 0.2 * 10 kg * 9.8 m/s² = 19.6 N

Work done by the applied force:

W_applied = Fd = (20 N)(4 m) = 80 J

Work done by friction:

W_friction = Ff * d * cos(180°) = -19.6 N * 4 m = -78.4 J (negative because friction opposes motion)

Problem 7: Power

A motor lifts a 200 kg object to a height of 10 meters in 5 seconds. Calculate the power output of the motor.

Solution:

First, calculate the work done by the motor:

Work (W) = mgh = 200 kg * 9.8 m/s² * 10 m = 19600 J

Power (P) is the rate at which work is done:

P = W / t = 19600 J / 5 s = 3920 W

The power output of the motor is 3920 Watts.

III. Explanation of Scientific Principles

These problems illustrate several key scientific principles:

• Conservation of Energy: In problems without friction or other non-conservative forces, the total mechanical energy (potential energy + kinetic energy) remains constant. Energy is transformed from one form to another, but the total amount remains the same.

• Work-Energy Theorem: This theorem provides a powerful link between work and kinetic energy. The net work done on an object is directly responsible for the change in its kinetic energy.

• Friction: Friction is a non-conservative force that opposes motion and converts mechanical energy into thermal energy (heat). The work done by friction is always negative.

IV. Frequently Asked Questions (FAQ)

• What is the difference between work and energy? Work is the process of transferring energy, while energy is the capacity to do work.

• Why is the work done by friction always negative? Friction opposes motion, so the force of friction is always in the opposite direction of the displacement. Since cos(180°) = -1, the work done by friction is negative.

• What happens to the energy lost due to friction? It is converted into thermal energy (heat), increasing the temperature of the surfaces in contact.

V. Conclusion

Understanding work and energy is crucial for solving a variety of physics and engineering problems. By applying the fundamental principles and equations discussed in this guide, and by working through numerous sample problems, you'll develop a strong foundation for tackling more complex scenarios in the future. Remember to carefully analyze each problem, identify the relevant forces and displacements, and use the appropriate equations to find the solution. Consistent practice is key to mastering these concepts.