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Chapter 5: Problem 25

A stockroom worker pushes a box with mass \(16.8 \mathrm{~kg}\) on a horizontal surface with a constant speed of \(3.50 \mathrm{~m} / \mathrm{s}\). 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
The horizontal force needed to maintain the motion of the box is found by \(F_a = \mu_k * m * g\). With the force removed, the distance the box moves before coming to rest can be found with \(d_b = (v^2 - u^2) / (2*a)\).

Step by step solution

01

Understand the Problem

Identify variables of the problem: the mass of the box \(m = 16.8\, \mathrm{kg}\), the speed of the box \(v = 3.50\, \mathrm{m/s}\), the coefficient of kinetic friction between the box and the surface \(\mu_k = 0.20\). Problem a) consists of finding the force \(F_a\) that needs to be exerted to keep the box moving at constant velocity. Once the force is found, b) consists of finding the distance \(d_b\) the box moves before coming to rest.
02

Calculate the force required to maintain the box's motion

Since the box moves at a constant speed, it means there is no acceleration. Therefore, by Newton's second law, the net force must be zero and the applied force \(F_a\) must balance the friction force \(F_{friction}=\mu_k * F_{normal}\). In this case, the normal force is equal to the weight of the box, so \(F_{normal} = m*g\), with \(g = 9.8 \, \mathrm{m/s^2}\). Hence, \(F_a = \mu_k * F_{normal} = \mu_k * m * g\).
03

Compute the distance the box moves before coming to rest

When the horizontal force is removed, only the friction force acts on the box, which is a retarding force. By second law of motion, \(F = m*a\), here \(F = F_{friction} = \mu_k * m * g\) and acceleration \(a = F/m = \mu_k * g \). Given the initial speed \(v\) and knowing that the final speed is zero, we can use the equation \(v^2 = u^2 + 2*a*d_b\) to solve for distance \(d_b\). Since the box comes to stop, \(v = 0\), so the equation can be rewritten as \(d_b = (v^2 - u^2) / 2*a\).
04

Final Calculation

By inserting the given values to the equations in Step 2 and Step 3, we can find out the values of \(F_a\) and \(d_b\).

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Key Concepts

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

Newton's Second Law of Motion
Understanding Newton's second law of motion is fundamental to analyzing most physical systems, including the motion of an object against friction. According to this law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The law is commonly expressed as the equation:
\[ F = ma \]
where \(F\) is the net force applied to the object, \(m\) is the mass of the object, and \(a\) is the acceleration.

In our case of a stockroom worker pushing a box across a surface, the law becomes particularly applicable. As the problem states that the box is moving with a constant velocity, we infer that the acceleration (\(a\)) is zero because constant velocity implies no change in speed or direction. This leads us to conclude that the net force is also zero, which means the pushing force exerted by the worker is equal and opposite to the force of friction. This is a direct application of Newton's second law.

When any additional force is removed, and only friction acts on the box, the second law tells us that the box will decelerate (negative acceleration) and eventually come to a stop. The amount of force due to friction, which causes this deceleration, can be calculated through the law as well.
Coefficient of Kinetic Friction
The coefficient of kinetic friction (\( \( \mu_k \) \)) is a dimensionless scalar value that describes the ratio of the force of friction between two bodies and the force pressing them together. It has no units because it is a ratio of two forces. The force of kinetic friction can be calculated using the equation:
\[ F_{friction} = \mu_k F_{normal} \]
where \(F_{friction}\) is the force of friction and \(F_{normal}\) is the normal force, which in this case equals to the gravitational force on the box (\(m*g\)).

In our textbook exercise, the coefficient of kinetic friction is a crucial factor in determining how much force the worker must exert and how far the box will slide after the force is removed. The smaller the coefficient, the less force is needed to maintain the box's motion, and the further it will slide when the force is removed. The constant value of 0.20 in this case shows us the resistance to motion the box faces on the surface.
Constant Velocity Motion
The scenario described in the problem demonstrates constant velocity motion, which is a key concept in kinematics. Constant velocity means that the object in question is moving in a straight line at an unchanging speed. Moreover, if velocity is constant, there is no acceleration (\(a = 0\)) since acceleration is defined as the rate of change of velocity.

Understanding this allows us to solve for the forces at play intuitively. For an object to move at constant velocity, the net force acting on the object must be zero; hence the forces acting on the object are in balance. In this context, the constant velocity tells us that the applied force by the worker is exactly balancing the kinetic frictional force.

Additionally, the concept is used when the force is removed. Even though the box will eventually come to a stop due to kinetic friction, the motion between the removal of the force and stopping is also a constant velocity motion, for each infinitesimal moment until the box stops, as the velocity decreases uniformly due to the constant frictional force.

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Most popular questions from this chapter

Block \(B,\) with mass \(5.00 \mathrm{~kg},\) rests on block \(A,\) with mass \(8.00 \mathrm{~kg}\), which in turn is on a horizontal tabletop (Fig. P5.92). There is no friction between block \(A\) and the tabletop, but the coefficient of static friction between blocks \(A\) and \(B\) is \(0.750 .\) A light string attached to block \(A\) passes over a friction less, massless pulley, and block \(C\) is suspended from the other end of the string. What is the largest mass that block \(C\) can have so that blocks \(A\) and \(B\) still slide together when the system is released from rest?

Force on a Skater's Wrist. A \(52 \mathrm{~kg}\) ice skater spins about a vertical axis through her body with her arms horizontally outstretched; she makes 2.0 turns each second. The distance from one hand to the other is \(1.50 \mathrm{~m} .\) Biometric measurements indicate that each hand typically makes up about \(1.25 \%\) of body weight. (a) Draw a free-body diagram of one of the skater's hands. (b) What horizontal force must her wrist exert on her hand? (c) Express the force in part (b) as a multiple of the weight of her hand.

A \(2540 \mathrm{~kg}\) test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by \(v(t)=A t+B t^{2}\), where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) at the instant of ignition and, 1.00 s later, an upward velocity of \(2.00 \mathrm{~m} / \mathrm{s}\). (a) Determine \(A\) and \(B\), including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

Runway Design. A transport plane takes off from a level landing field with two gliders in tow, one behind the other. The mass of each glider is \(700 \mathrm{~kg}\), and the total resistance (air drag plus friction with the runway on each may be assumed constant and equal to \(2500 \mathrm{~N}\). The tension in the towrope between the transport plane and the first glider is not to exceed \(12,000 \mathrm{~N}\). (a) If a speed of \(40 \mathrm{~m} / \mathrm{s}\) is required for takeoff, what minimum length of runway is needed? (b) What is the tension in the towrope between the two gliders while they are accelerating for the takeoff?

Jack sits in the chair of a Ferris wheel that is rotating at a constant \(0.100 \mathrm{rev} / \mathrm{s}\). As Jack passes through the highest point of his circular path, the upward force that the chair exerts on him is equal to one- fourth of his weight. What is the radius of the circle in which Jack travels? Treat him as a point mass.
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