Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A sled with mass 12.00 kg moves in a straight line on a frictionless, horizontal surface. At one point in its path, its speed is 4.00 m/s; after it has traveled 2.50 m beyond this point, its speed is 6.00 m/s. Use the work\(-\)energy theorem to find the force acting on the sled, assuming that this force is constant and that it acts in the direction of the sled's motion.

Short Answer

Expert verified
The constant force acting on the sled is 48.00 N.

Step by step solution

01

Identify the Known Parameters

We need to gather all the known quantities from the problem:1. Initial velocity of the sled, \( v_i = 4.00 \, \text{m/s} \).2. Final velocity of the sled, \( v_f = 6.00 \, \text{m/s} \).3. Distance traveled, \( d = 2.50 \, \text{m} \).4. Mass of the sled, \( m = 12.00 \, \text{kg} \).
02

Apply the Work-Energy Theorem

The work-energy theorem states that the work done on the sled is equal to the change in its kinetic energy. The expression for work done \( W \) is given by:\[ W = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \].Substituting the known values:\[ W = \frac{1}{2} \times 12.00 \, \text{kg} \times (6.00 \, \text{m/s})^2 - \frac{1}{2} \times 12.00 \, \text{kg} \times (4.00 \, \text{m/s})^2 \].
03

Calculate the Change in Kinetic Energy

First, calculate the initial kinetic energy:\[ KE_i = \frac{1}{2} \times 12.00 \times (4.00)^2 = 96.00 \, \text{J} \].Next, calculate the final kinetic energy:\[ KE_f = \frac{1}{2} \times 12.00 \times (6.00)^2 = 216.00 \, \text{J} \].Now, find the change in kinetic energy:\[ \Delta KE = 216.00 \, \text{J} - 96.00 \, \text{J} = 120.00 \, \text{J} \].
04

Relate Work Done to Force

The work done \( W \) is also given by \( W = F \cdot d \), where \( F \) is the force and \( d \) is the distance. We know that \( W = 120.00 \, \text{J} \) and \( d = 2.50 \, \text{m} \). Substituting these into the equation:\[ 120.00 \, \text{J} = F \times 2.50 \, \text{m} \].Solve for \( F \):\[ F = \frac{120.00 \, \text{J}}{2.50 \, \text{m}} = 48.00 \, \text{N} \].
05

Conclusion

The constant force acting on the sled, which causes its speed to increase from 4.00 m/s to 6.00 m/s over a distance of 2.50 m, is 48.00 N.

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 Vaia!

Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. In our scenario, this energy is crucial in understanding how the sled moves and how forces affect it. The formula for kinetic energy (KE) is given by: \[ KE = \frac{1}{2} mv^2 \] where \( m \) is the mass of the object and \( v \) is its velocity.
Applying this formula to the sled, we calculate its initial kinetic energy at a speed of 4.00 m/s and its final kinetic energy at 6.00 m/s.
  • Initial kinetic energy: \( KE_i = \frac{1}{2} \times 12.00 \times (4.00)^2 = 96.00 \text{ J} \)
  • Final kinetic energy: \( KE_f = \frac{1}{2} \times 12.00 \times (6.00)^2 = 216.00 \text{ J} \)
The change in kinetic energy helps in determining the work done on the sled. Understanding how kinetic energy changes reveals the impact of forces on the sled's motion.
The Role of Constant Force
A constant force is one that remains the same in magnitude and direction throughout the problem. In the context of this exercise, we are tasked with finding the constant force that accelerates the sled from 4.00 m/s to 6.00 m/s.
This constant force acts on the sled over a distance of 2.50 meters. According to the work-energy theorem, the work done by this force equals the change in kinetic energy.
  • Total work done \( W = 120.00 \text{ J} \), which is calculated from the change in kinetic energy.
We use the formula for work \( W \), which is the product of force \( F \) and distance \( d \):\[ W = F \cdot d \]By rearranging to solve for \( F \), we find:\[ F = \frac{W}{d} = \frac{120.00 \text{ J}}{2.50 \text{ m}} = 48.00 \text{ N} \]This reveals the constant force responsible for the sled's acceleration.
Exploring Sled Motion
Sled motion can be influenced by various factors and forces. In a physics problem, understanding the motion of a sled involves analyzing how different forces come into play on a given path.
In this exercise, the sled moves in a straight line and undergoes an increase in speed due to a constant force exerted over a certain distance.
Key elements of sled motion include:
  • Direction of motion: The force is applied in the same direction as the sled's movement, making it easier to calculate its effect on speed.
  • Acceleration: The increase in speed from 4.00 m/s to 6.00 m/s signifies acceleration, which is a change in velocity over time.
Sled motion in this frictionless scenario also relates back to the sled's kinetic energy, providing a deeper understanding of how forces and motion interrelate in physics.
The Implications of a Frictionless Surface
When considering sled motion, the presence of friction can significantly alter the dynamics of a situation. However, this exercise takes place on a frictionless surface, simplifying the problem.
On a frictionless surface:
  • There is no opposing force to slow the sled down. This focus allows us to concentrate on the effects of the constant force.
  • The only force that influences the sled’s acceleration is the external constant force applied.
The absence of friction allows for a clear application of Newton's laws. Frictionless surfaces are ideal for learning fundamental physics concepts because they eliminate unnecessary complexities, helping students understand core mechanics and motion principles without the additional layer of resistive forces complicating calculations.

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

Two blocks are connected by a very light string passing over a massless and frictionless pulley (\(\textbf{Fig. E6.7}\)). Traveling at constant speed, the 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward. How much work is done (a) on the 12.0-N block by (i) gravity and (ii) the tension in the string? (b) How much work is done on the 20.0-N block by (i) gravity, (ii) the tension in the string, (iii) friction, and (iv) the normal force? (c) Find the total work done on each block.

A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0\(^\circ\) above the horizontal by a rope exerting a 72.0-N pull parallel to the ramp's surface. If the carton travels 5.20 m along the surface of the ramp, calculate the work done on it by (a) the rope, (b) gravity, and (c) the normal force of the ramp. (d) What is the net work done on the carton? (e) Suppose that the rope is angled at 50.0\(^\circ\) above the horizontal, instead of being parallel to the ramp's surface. How much work does the rope do on the carton in this case?

A pump is required to lift 800 kg of water (about 210 gallons) per minute from a well 14.0 m deep and eject it with a speed of 18.0 m/s. (a) How much work is done per minute in lifting the water? (b) How much work is done in giving the water the kinetic energy it has when ejected? (c) What must be the power output of the pump?

A block of ice with mass 2.00 kg slides 1.35 m down an inclined plane that slopes downward at an angle of 36.9\(^\circ\) below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.

A soccer ball with mass 0.420 kg is initially moving with speed 2.00 m/s. A soccer player kicks the ball, exerting a constant force of magnitude 40.0 N in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 m/s?

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.

Sign-up for free