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Chapter 5: Problem 27

A stockroom worker pushes a box with mass 16.8 kg on a horizontal surface with a constant speed of 3.50 m/s. The coefficient of kinetic friction between the box and the surface is 0.20. (a) What horizontal force must the worker apply to maintain the motion? (b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest?

Short Answer

Expert verified
(a) The horizontal force is 32.95 N. (b) The box slides 3.13 m before stopping.

Step by step solution

01

Understand the Problem

The problem involves calculating the horizontal force needed to maintain the constant speed of the box on a surface with friction, and also determining how far the box will slide if this force is removed. This requires understanding concepts like force, friction, speed, and distance.
02

Identify Given Values

From the problem, we identify:- Mass of the box, \( m = 16.8 \, \text{kg} \)- Speed of the box, \( v = 3.50 \, \text{m/s} \)- Coefficient of kinetic friction, \( \mu_k = 0.20 \)- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \) (assumed).
03

Calculate the Frictional Force

The kinetic frictional force, \( f_k \), can be calculated using the formula:\[ f_k = \mu_k \cdot m \cdot g \]Substitute the given values:\[ f_k = 0.20 \cdot 16.8 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 32.95 \, \text{N} \]
04

Determine the Required Horizontal Force

Since the box is moving at a constant speed, the horizontal force exerted by the worker, \( F \), must equal the frictional force. Therefore:\[ F = f_k = 32.95 \, \text{N} \]
05

Calculate the Deceleration

Once the worker stops applying the force, the only force acting horizontally is friction, which provides deceleration. Use the formula for acceleration:\[ a = \frac{f_k}{m} \]\[ a = \frac{32.95 \, \text{N}}{16.8 \, \text{kg}} = 1.96 \, \text{m/s}^2 \]
06

Compute Distance the Box Slides

Use the equation \( v_f^2 = v_i^2 + 2a d \) to find the distance \( d \):- Initial velocity \( v_i = 3.50 \, \text{m/s} \)- Final velocity \( v_f = 0 \, \text{m/s} \)- The deceleration \( a = -1.96 \, \text{m/s}^2 \)Rearrange to solve for \( d \):\[ 0 = (3.50)^2 + 2(-1.96)d \]\[ d = \frac{(3.50)^2}{2 \times 1.96} = 3.13 \, \text{m} \]
07

Record the Final Answers

For part (a), the horizontal force required is \( 32.95 \, \text{N} \). For part (b), the box slides \( 3.13 \, \text{m} \) before coming to rest.

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Key Concepts

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

Constant Speed
When an object moves at a constant speed, it means that the velocity of the object never changes over time. In the context of our exercise, a box is being pushed by a worker at a constant speed of 3.50 m/s. This indicates that the forces acting on the box are perfectly balanced.

There are two primary forces to consider here:
  • The force exerted by the worker pushing the box.
  • The kinetic frictional force opposing the motion.
For the speed to remain constant, the horizontal force applied by the worker must exactly counteract the frictional force. Thus, there is no net force acting on the box, ensuring it doesn't accelerate or decelerate. Understanding this balance of forces is crucial when dealing with problems involving constant speed.
Horizontal Force
To maintain the box's constant speed against friction, the worker must apply a precise amount of horizontal force. This force directly matches the kinetic frictional force acting in the opposite direction.

The formula to calculate the kinetic frictional force is:
  • \( f_k = \, \mu_k \times m \times g \)
Where:
  • \( \mu_k \) is the coefficient of kinetic friction (0.20 in this exercise)
  • \( m \) is the mass of the box (16.8 kg)
  • \( g \) is the acceleration due to gravity (9.81 m/s虏)
The calculated force is 32.95 N, which is the exact force the worker needs to apply horizontally. This knowledge allows us to understand how frictional resistance is overcome to keep an object moving steadily.
Deceleration
Deceleration refers to the slowing down of an object. It occurs when an object experiences a force in the opposite direction of its motion. In this exercise, once the worker stops pushing the box, the only remaining force acting on it is friction.

The kinetic frictional force becomes the decelerating force, slowing the box down.
The formula for calculating deceleration given by friction is:
  • \( a = \frac{f_k}{m} \)
For the box:
  • \( f_k = 32.95 \, \text{N} \)
  • \( m = 16.8 \, \text{kg} \)
  • This results in a deceleration of 1.96 m/s虏.
Deceleration implies that the velocity of the box decreases until it eventually comes to a stop. Understanding how deceleration works help explain why an object must come to rest when external forces are no longer applied.
Sliding Distance
The sliding distance refers to how far the box travels after the worker stops applying the force. As the only acting force now is the friction, it will decelerate the box to a stop.

The initial velocity of the box is 3.50 m/s, and we use the formula:
  • \( v_f^2 = v_i^2 + 2a d \)
to find the sliding distance \( d \).
  • \( v_f \) is the final velocity (0 m/s, as the box stops).
  • \( v_i \) is the initial velocity (3.50 m/s).
  • \( a \) is the deceleration (-1.96 m/s虏).
By solving, \( d \) is found to be 3.13 meters. This represents how far the box travels due to its inertia before coming to a complete stop. This calculation showcases the effects of force removal and friction on an object's motion over a distance.

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Most popular questions from this chapter

A 5.00-kg box sits at rest at the bottom of a ramp that is 8.00 m long and is inclined at 30.0\(^\circ\) above the horizontal. The coefficient of kinetic friction is \(\mu_k\) = 0.40, and the coefficient of static friction is \(\mu_s\) = 0.43. What constant force \(F\), applied parallel to the surface of the ramp, is required to push the box to the top of the ramp in a time of 6.00 s?

A 25.0-kg box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is 0.25, and the coefficient of static friction is 0.35. (a) As \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 m along the loading ramp?

A 5.00-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force \(F(t)\) is applied to the end of the rope, and the height of the crate above its initial position is given by \(y(t) =\) (2.80 m/s)\(t +\) (0.610 m/s\(^3\))\(t^3\). What is the magnitude of F when \(t =\) 4.00 s?

A 1125-kg car and a 2250-kg pickup truck approach a curve on a highway that has a radius of 225 m. (a) At what angle should the highway engineer bank this curve so that vehicles traveling at 65.0 mi/h can safely round it regardless of the condition of their tires? Should the heavy truck go slower than the lighter car? (b) As the car and truck round the curve at 65.0 mi/h, find the normal force on each one due to the highway surface.

You throw a baseball straight upward. The drag force is proportional to \(v^2\). In terms of \(g\), what is the y-component of the ball's acceleration when the ball's speed is half its terminal speed and (a) it is moving up? (b) It is moving back down?
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