/*! 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 6 A horizontal force of \(150 \mat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A horizontal force of \(150 \mathrm{~N}\) is used to push a \(40.0-\mathrm{kg}\) packing crate a distance of \(6.00 \mathrm{~m}\) on a rough horizontal surface. If the crate moves at constant speed, find (a) the work done by the \(150-\mathrm{N}\) force and (b) the coefficient of kinetic friction between the crate and surface.

Short Answer

Expert verified
(a) The work done by the 150-N force is \(900 J\). (b) The coefficient of kinetic friction is approximately 0.382.

Step by step solution

01

Calculate the Work Done

The work done by the 150-N force is calculated using the formula \(W = F \cdot d \cdot \cos(\theta)\), where \(F\) is force, \(d\) is displacement, and \(\theta\) is the angle between the force and displacement. In this case, the force and displacement are in the same direction (horizontally), so \(\theta = 0\). Therefore, \(W = 150N \cdot 6.00m \cdot \cos(0º)\). Calculating this will give us the work done.
02

Find the friction force

Since the object is moving at a constant speed, the net force on the object is zero (from Newton's second law of motion). This means the applied force is equal to the friction force, i.e., \(F_{applied} = F_{friction}\). So, the friction force equals 150N.
03

Find the normal force

The normal force is the force exerted by a surface that supports the weight of an object resting on it. It acts perpendicular (or 'normal') to the surface. The normal force is equal to the weight of the object, which is its mass multiplied by gravity: \(F_{normal} = m \cdot g\), where \(m = 40.0 kg\) and \(g = 9.8 m/s^2\). Calculate \(F_{normal}\) for further use.
04

Calculate the Coefficient of kinetic friction

The coefficient of kinetic friction is given by \(μ_k = F_{friction} / F_{normal}\). We have calculated both \(F_{friction}\) and \(F_{normal}\) in the previous steps. So, plug in the values and calculate \(\mu_k\) for the final answer.

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.

Newton's Laws of Motion
When discussing motion, Newton's Laws are fundamental to understanding how objects behave either at rest or in motion. Newton's First Law, also known as the law of inertia, states that an object will remain at rest or move at a constant speed in a straight line unless acted upon by an external force. This concept applies to our crate problem because the crate is moving at a constant speed. This implies that the total force acting on it is zero.

Newton's Second Law gives us the relationship between force, mass, and acceleration, expressed by the equation \(F = m \cdot a\). When the crate moves at constant speed, the acceleration \(a\) is zero, hence the net force is zero. Thus, the force exerted by the pusher must equal the frictional force opposing the crate's motion.
  • This balance ensures that the forces are equal and opposite.
  • Knowing this allows us to equate the applied force to the frictional force.
Coefficient of Kinetic Friction
The coefficient of kinetic friction, represented by \( \mu_k \), is a dimensionless constant that characterizes the amount of friction between two objects in motion. It varies based on the materials in contact and is vital for calculating frictional forces.

In the given problem, the kinetic frictional force must counteract the applied force since the crate moves at a constant speed, as dictated by Newton's laws. This means \( F_{applied} = F_{friction} \). The coefficient of kinetic friction is found using the formula \( \mu_k = \frac{F_{friction}}{F_{normal}} \).
  • The normal force is equivalent to the weight of the crate, i.e., mass \( \times \) gravity.
  • Substitute known values into the mentioned formula to find \( \mu_k \).
  • Understanding \( \mu_k \) helps us predict how easily the crate will slide over the surface.
Frictional Force
Friction is a force that resists the relative motion of two surfaces in contact. It's crucial to understanding our crate’s motion along the surface. In this problem, we deal with kinetic friction, which operates when two objects slide over each other.

The frictional force depends on:
  • The coefficient of kinetic friction \( \mu_k \).
  • The normal force \( F_{normal} \).
This relationship is expressed with the equation \( F_{friction} = \mu_k \cdot F_{normal} \).

Since the applied force equals the frictional force (due to constant speed), calculating \( F_{friction} \) helps us find \( \mu_k \), and subsequently, further emphasizes the role of friction in everyday physics interactions.
Constant Speed Motion
Motion at a constant speed refers to an object moving at a uniform rate, without speeding up or slowing down. In this scenario, the velocity of the crate remains unchanged while it is being pushed across a rough surface.

Several critical conditions are necessary:
  • The forces applied to the crate must balance.
  • The applied force must equal the frictional force.
Achieving constant speed means the net force is zero, ensuring no acceleration. This is vital since it leads us to use the equality \( F_{applied} = F_{friction} \), allowing us to calculate the coefficient of kinetic friction effortlessly.

Understanding constant speed motion clarifies why the energy input through pushing is precisely matched by the energy lost to friction, offering an excellent scenario to apply principles of work and energy.

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 \(1.50 \times 10^{3}\) -kg car starts from rest and accelerates uniformly to \(18.0 \mathrm{~m} / \mathrm{s}\) in \(12.0 \mathrm{~s}\). Assume that air resistance remains constant at \(400 \mathrm{~N}\) during this time. Find (a) the average power developed by the engine and (b) the instantaneous power output of the engine at \(t=12.0 \mathrm{~s}\), just before the car stops accelerating.

A block of mass \(12.0 \mathrm{~kg}\) slides from rest down a frictionless \(35,0^{\circ}\) incline and is stopped by a strong spring with \(k=3.00 \times 10^{4} \mathrm{~N} / \mathrm{m} .\) The block slides \(3.00 \mathrm{~m}\) from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed?

An archer pulls her bowstring back \(0.400 \mathrm{~m}\) by exerting a force that increases uniformly from zero to \(230 \mathrm{~N}\). (a) What is the equivalent spring constant of the bow? (b) How much work does the archer do in pulling the bow?

While running, a person dissipates about \(0.60 \mathrm{~J}\) of mechanical energy per step per kilogram of body mass. If a \(60-\mathrm{kg}\) person develops a power of \(70 \mathrm{~W}\) during a race, how fast is the person running? (Assume a running step is \(1.5 \mathrm{~m}\) long.)

A shopper in a supermarket pushes a cart with a force of \(95 \mathrm{~N}\) directed at an angle of \(25^{\circ}\) below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at constant speed. (a) Find the work done by the shopper as she moves down a \(50.0-\mathrm{m}\) length aisle, (b) What is the net work done on the cart? Why? (c) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesn't change, would the shopper's applied force be larger, smaller, or the same? What about the work done on the cart by the shopper?

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.