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

You push a large crate across the floor at constant speed, exerting a horizontal force \(F\) on the crate. There is friction between the floor and the crate. The force of friction has a magnitude that is a) zero. b) \(F\). c) greater than \(F\). information. d) less than \(F\). e) impossible to quantify without further

Short Answer

Expert verified
Answer: b) \(F\).

Step by step solution

01

Newton's second law

According to Newton's second law, the net force acting on an object equals the mass of the object multiplied by its acceleration (\(F_{net} = m*a\)). Since the crate is moving at a constant speed, its acceleration is 0.
02

Friction force

The friction force is the force opposing the motion of the object. It acts in the opposite direction of the applied force \(F\). The friction force can be calculated using the formula: \(F_{friction} = μ * F_{normal}\), where \(μ\) is the coefficient of friction, and \(F_{normal}\) is the normal force acting perpendicular to the surface (in this case, the weight of the crate).
03

Equilibrium condition

As the crate is moving at a constant speed, the net force acting on it should be 0. This means that the applied force \(F\) and the friction force \(F_{friction}\) should balance each other out. Therefore, \(F_{net} = F - F_{friction} = 0\), which implies that \(F = F_{friction}\). Thus, the correct answer is: b) \(F\).

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.

Force of Friction
Understanding the force of friction is crucial when dealing with motion and physics problems. It's the resistive force that opposes an object's motion across a surface.

When you push an object like a crate across the floor, friction acts in the opposite direction of your push. This force is not random; it directly relates to the nature of the surfaces in contact and the force pressing the surfaces together, which is referred to as the normal force.

One common misconception with friction is that it increases with speed. However, the frictional force is independent of the speed as long as the conditions of the contact surfaces and the normal force remain constant. Instead, friction depends on the types of surfaces and the coefficient of friction—a measure of how 'sticky' or 'slippery' a surface is.
Constant Speed
Moving at a constant speed means that the velocity of an object is not changing. Velocity encompasses both speed and direction. Thus, constant velocity implies no acceleration or deceleration is occurring—speed remains steady and the direction unaltered.

In scenarios like pushing a crate across a floor, if the crate moves at a constant speed despite your continuous push, this indicates that all forces acting on the crate are balanced. No net force is present to change the state of motion of the crate.

Principal to understanding constant velocity is Newton's first law of motion, which suggests that an object will remain at rest or in uniform motion unless acted upon by an external force. Here, constant speed implies that, after you start pushing, you don't need to increase or decrease your effort to maintain the crate's motion, assuming friction and other conditions remain constant.
Equilibrium Condition
In physics, when all forces acting upon an object are balanced, the object is said to be in a state of equilibrium. There are two types of equilibrium: static, where the object is at rest, and dynamic, where the object moves with a constant velocity.

Dynamic equilibrium pertains to our example with the crate. As you apply a horizontal force to move the crate at a constant speed, the force of friction equally opposes this applied force. In such a balanced state, the net force on the crate is zero, meeting the equilibrium condition. This condition is at the heart of Newton’s first law of motion, emphasizing that an object in motion will not change its velocity unless a net force is applied to it.
Net Force
The net force is the vector sum of all the forces acting on an object. It defines the overall impact of all those forces on the object's state of motion.

If an object experiences several forces, they can either add up constructively or destructively, depending on their directions. In the exercise with the crate, the horizontal pushing force and the opposing force of friction are the primary forces at play. If the crate is moving at a constant speed, it tells us that these forces are equal in magnitude and opposite in direction—thus, their sum, the net force, is zero.

A net force of zero is profound because it directly implies that the object will either remain at rest or move at a constant velocity, adhering to the concept of dynamic equilibrium covered in Newton’s laws of motion.
Acceleration
Acceleration is the rate at which an object’s velocity changes over time. It can involve an increase or decrease in speed and/or a change in direction.

In physics equations, acceleration is expressed as 'a' and is typically measured in meters per second squared (m/s²). Newton's second law of motion connects force, mass, and acceleration, stating that the net force on an object is equal to its mass times its acceleration (\(F_{net} = m \times a\)).

Remember, when a body moves at a constant velocity, like the crate in the question, its acceleration is zero since its speed doesn't change. In our scenario, this means that no additional net force is required for the crate to keep moving at that steady speed.

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 piñata of mass \(M=12.0 \mathrm{~kg}\) hangs on a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is \(D=2.00 \mathrm{~m},\) the top of the right pole is a vertical distance \(h=0.500 \mathrm{~m}\) higher than the top of the left pole, and the total length of the rope between the poles is \(L=3.00 \mathrm{~m} .\) The piñata is attached to a ring, with the rope passing through the center of the ring. The ring is frictionless, so that it can slide freely along the rope until the piñata comes to a point of static equilibrium. a) Determine the distance from the top of the left (lower) pole to the ring when the pinata is in static equilibrium. b) What is the tension in the rope when the pinata is at this point of static equilibrium?

A tractor pulls a sled of mass \(M=1000\). kg across level ground. The coefficient of kinetic friction between the sled and the ground is \(\mu_{\mathrm{k}}=0.600 .\) The tractor pulls the sled by a rope that connects to the sled at an angle of \(\theta=30.0^{\circ}\) above the horizontal. What magnitude of tension in the rope is necessary to move the sled horizontally with an acceleration \(a=2.00 \mathrm{~m} / \mathrm{s}^{2}\) ?

A hanging mass, \(M_{1}=0.400 \mathrm{~kg}\), is attached by a light string that runs over a frictionless pulley to a mass \(M_{2}=1.20 \mathrm{~kg}\) that is initially at rest on a frictionless ramp. The ramp is at an angle of \(\theta=30.0^{\circ}\) above the horizontal, and the pulley is at the top of the ramp. Find the magnitude and direction of the acceleration, \(a_{2},\) of \(M_{2}\)

A load of bricks of mass \(M=200.0 \mathrm{~kg}\) is attached to a crane by a cable of negligible mass and length \(L=3.00 \mathrm{~m}\). Initially, when the cable hangs vertically downward, the bricks are a horizontal distance \(D=1.50 \mathrm{~m}\) from the wall where the bricks are to be placed. What is the magnitude of the horizontal force that must be applied to the load of bricks (without moving the crane) so that the bricks will rest directly above the wall?

A skydiver of mass \(82.3 \mathrm{~kg}\) (including outfit and equipment) floats downward suspended from her parachute, having reached terminal speed. The drag coefficient is 0.533 , and the area of her parachute is \(20.11 \mathrm{~m}^{2}\). The density of air is \(1.14 \mathrm{~kg} / \mathrm{m}^{3}\). What is the air's drag force on her?

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