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Chapter 5: Problem 39

\(\bullet$$\bullet\)A stockroom worker pushes a box with mass 11.2 \(\mathrm{kg}\) on a horizontal surface with a constant speed of 3.50 \(\mathrm{m} / \mathrm{s}\) . The coef- ficients of kinetic and static friction between the box and the surface are 0.200 and \(0.450,\) respectively. (a) What horizontal force must the worker apply to maintain the motion of the box? (b) If the worker stops pushing, what will be the acceler- ation of the box?

Short Answer

Expert verified
(a) 21.97 N (b) -1.96 m/s虏

Step by step solution

01

Identify Given Values for Part (a)

We have a mass of the box, \(m = 11.2\, \text{kg}\), and the coefficient of kinetic friction, \(\mu_k = 0.200\). We need to find the horizontal force applied by the worker to maintain constant speed. Note that the gravitational force \(F_g\) acting on the box is \(m \cdot g\), where \(g = 9.81\, \text{m/s}^2\).
02

Calculate the Normal Force

The normal force \(N\) is equal to the gravitational force on the box since it is moving on a horizontal surface without any vertical acceleration. So, \(N = m \cdot g = 11.2\, \text{kg} \times 9.81\, \text{m/s}^2 = 109.872\, \text{N}\).
03

Calculate the Force of Kinetic Friction

The force of kinetic friction \(f_k\) can be calculated using the formula: \(f_k = \mu_k \times N\). So, \(f_k = 0.200 \times 109.872\, \text{N} = 21.9744\, \text{N}\).
04

Determine the Horizontal Force Required

To keep the box moving at a constant speed, the worker must apply a force equal to the force of kinetic friction. Therefore, the horizontal force required is \(F = f_k = 21.9744\, \text{N}\).
05

Identify Given Values for Part (b)

Now, for part (b), we find the box's acceleration when the worker stops pushing. We still have the mass, \(m = 11.2\, \text{kg}\), and the coefficient of kinetic friction, \(\mu_k = 0.200\).
06

Calculate the Acceleration of the Box

Once the worker stops pushing, the only horizontal force acting on the box is the kinetic friction, which will decelerate it. We use Newton's second law: \(F_{net} = m \cdot a\). The net force here is \(-f_k\) (since it's opposite to the motion direction), giving us \(-21.9744\, \text{N} = 11.2\, \text{kg} \cdot a\). Solving for acceleration: \(a = \frac{-21.9744\, \text{N}}{11.2\, \text{kg}} = -1.962\, \text{m/s}^2\).

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Key Concepts

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

Newton's Second Law
At the heart of understanding the forces at play in this problem is Newton's Second Law. This fundamental principle of physics tells us how an object will behave when it is under the influence of a force. In simple terms, the law states that the acceleration (abla a) of an object is directly proportional to the net force (abla F_{net}) acting on the object and inversely proportional to its mass (abla m). Mathematically, it's written as:
\[ F_{net} = m imes a \]It means that if you push something with a certain force, that object will accelerate based on its mass. The larger the mass, the smaller the acceleration for a given force, and vice-versa. This is exactly what we used in the problem to find how the box decelerates when the worker stops pushing.
Constant Speed Motion
Constant speed motion is a key concept here because it simplifies our calculations significantly. When an object moves at a constant speed, the net force acting on it is zero. This is because any force applied to the object is balanced by another, opposing force.
In the exercise, the worker needs to apply a force that matches the kinetic friction exactly to maintain a constant speed of the box. Since there is no acceleration, this means:
  • The force the worker applies ( abla F_{applied} ) equals the force of kinetic friction ( abla f_k ).
  • This balance of forces is what enables the box to continue its motion without speeding up or slowing down.
Therefore, understanding constant speed motion helps us know why the worker's force must precisely equal the frictional force.
Friction Coefficients
Friction is an important force in physics, and the friction coefficients are what quantify how much friction is present between two surfaces. Friction can act to impede motion, and it comes in two main types: static and kinetic. In this exercise, we focused on kinetic friction because the box was already moving.
The kinetic friction coefficient (abla \mu_k) tells us how much frictional force acts against the motion of the box. The formula:
\[ f_k = \mu_k \times N \]shows how to calculate this force, where (abla N) is the normal force acting on the object due to gravity.
The static friction coefficient is usually higher than the kinetic coefficient, meaning it is harder to start moving an object than it is to keep it moving once it's already in motion. This exercise asked for the force needed to keep the box at a constant speed, thus involving kinetic friction exclusively. Understanding these coefficients helps explain the resistance the worker had to overcome to keep the box moving.

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Most popular questions from this chapter

\(\bullet$$\bullet\) A winch is used to drag a 375 \(\mathrm{N}\) crate up a ramp at a con- stant speed of 75 \(\mathrm{cm} / \mathrm{s}\) by means of a rope that pulls parallel to the surface of the ramp. The rope slopes upward at \(33^{\circ}\) above the horizontal, and the coefficient of kinetic friction between the ramp and the crate is 0.25 . (a) What is the tension in the rope? (b) If the rope were suddenly to snap, what would be the accel- eration of the crate immediately after the rope broke?

\(\bullet$$\bullet\) A hockey puck leaves a player's stick with a speed of 9.9 \(\mathrm{m} / \mathrm{s}\) and slides 32.0 \(\mathrm{m}\) before coming to rest. Find the coef- ficient of friction between the puck and the ice.

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\(\bullet$$\bullet\) A \(25.0-\mathrm{kg}\) 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 the angle \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the accel- eration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 \(\mathrm{m}\) along the loading ramp"?

\(\bullet$$\bullet\) Forces during chin-ups. People who do chin-ups raise their chin just over a bar (the chinning bar), supporting themselves only by their arms. Typically, the body below the arms is raised by about 30 \(\mathrm{cm}\) in a time of 1.0 s, starting from rest. Assume that the entire body of a 680 N person who is chinning is raised this distance and that half the 1.0 \(\mathrm{s}\) is spent accelerating upward and the other half accelerating downward, uniformly in both cases. Make a free-body dia- gram of the person's body, and then apply it to find the force his arms must exert on him during the accelerating part of the chin-up.
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