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Chapter 5: Problem 37

\(\bullet$$\bullet\) An \(85-N\) box of oranges is being pushed across a horizon- tal floor. As it moves, it is slowing at a constant rate of 0.90 \(\mathrm{m} / \mathrm{s}\) each second. The push force has a horizontal compo- nent of 20 \(\mathrm{N}\) and a vertical component of 25 \(\mathrm{N}\) downward. Calculate the coefficient of kinetic friction between the box and floor.

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.25.

Step by step solution

01

Understand the Forces Acting on the Box

To solve the problem, we first need to understand the forces acting on the box. These include the gravitational force (weight), the horizontal push force, the vertical push force, and friction. The weight of the box provides a downward force of 85 N, while the vertical push also adds 25 N downward. The combined force is the normal force we need for calculating friction.
02

Calculate the Net Force Acting on the Box

The box is slowing down at a constant rate (decelerating), which means there is a net force acting in the opposite direction to the motion due to friction and less than the push force. \( F_{net} = ma \) where \( a = -0.90 \,\text{m/s}^2 \). To find mass, use \( m = \frac{85}{9.81} \approx 8.67 \, \text{kg} \). Hence, \( F_{net} = 8.67 \times (-0.90) \approx -7.80 \text{ N} \).
03

Calculate the Normal Force

The normal force \( N \) is the net force perpendicular to the surface. Considering the forces in the vertical direction, we have: \( N = 85 \, \text{N} + 25 \, \text{N} = 110 \, \text{N} \).
04

Calculate Frictional Force Required

The force of kinetic friction \( F_k \) opposes the motion and can be found by subtracting the net force from the horizontal component of the applied force. Hence, \( F_k = 20 \, \text{N} - (-7.80 \, \text{N}) = 27.80 \, \text{N} \).
05

Decide on Formula and Solve for Coefficient of Friction

The kinetic friction force is related to the normal force by the coefficient of friction: \( F_k = \mu_k N \). Therefore, the coefficient of kinetic friction \( \mu_k \) is \( \mu_k = \frac{27.80}{110} \approx 0.2536 \).

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Key Concepts

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

Coefficient of Friction
The coefficient of friction is a value that represents the amount of friction between two surfaces. In this problem, we are concerned with kinetic friction, which occurs when the box of oranges slides across the floor.
The coefficient of friction is a unitless number, usually between 0 and 1. It indicates how much frictional force is present per unit of normal force.
For our box, it's calculated using the formula: \[\mu_k = \frac{F_k}{N}\]- Here, \( F_k \) is the frictional force, already found to be 27.80 N.- \( N \) is the normal force and has been calculated as 110 N.By plugging in these values, the coefficient of kinetic friction is calculated as approximately 0.2536. This means that 25.36% of the normal force is contributing to friction. This value helps us understand how smoothly or roughly the box moves across the floor. A higher value would mean more friction, while lower indicates less resistance.
Normal Force
The normal force is the support force exerted by a surface, acting perpendicular to the surface itself. For the box of oranges, the floor exerts this force upwards, balancing the forces acting on the box.
In this scenario, it's not just the weight of the box contributing to the normal force; we must also consider the additional vertical push.
- The weight of the box exerts a 85 N force downward. - Additionally, a 25 N vertical force also pushes downward. These forces add up to give a total normal force of 110 N. Therefore, the normal force is greater than just the box's weight due to the extra downward push. This extra force makes the friction more significant than it would be with just the box alone.
Net Force
The net force is the vector sum of all forces acting on an object. It is what causes the object to accelerate or decelerate. In this exercise, the net force is what causes the box to slow down.
- The box is decelerating at a rate of 0.90 m/s虏.- Using Newton's second law, \( F_{net} = ma \), we determine the net force.The mass of the box is calculated by dividing its weight by gravity: \[m = \frac{85}{9.81} \approx 8.67\, \text{kg} \]Thus, the net force acting in the horizontal direction is approximately -7.80 N. This force opposes the direction of push due to friction, causing the box to slow down.
Newton's Laws
Newton's laws of motion are key to understanding how forces affect motion. They provide a framework for analyzing the forces acting on the box.

Newton's First Law

An object will stay at rest or move in a straight line at constant speed unless acted upon by an external force. In this case, friction acts as the external force slowing the box down.

Newton's Second Law

It tells us that the acceleration of an object depends on the net force acting upon it and its mass, summarized by \( F = ma \). We use this law here to relate the box's deceleration to the frictional force and the total applied force.

Newton's Third Law

This states that for every action, there is an equal and opposite reaction. While not explicitly calculated in the problem, it's reflected in how the floor pushes back with the normal force in response to the box's weight and the additional push downwards. By applying these laws, we can comprehensively analyze the dynamics of the box as it slides across the floor.

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Most popular questions from this chapter

\(\bullet\) An unstretched spring is 12.00 \(\mathrm{cm}\) long. When you hang an 875 g weight from it, it stretches to a length of 14.40 \(\mathrm{cm}\) . (a) What is the force constant (in \(\mathrm{N} / \mathrm{m}\) ) of this spring? (b) What total mass must you hang from the spring to stretch it to a total length of 17.72 \(\mathrm{cm} ?\)

\(\bullet$$\bullet\) A crate of 45.0 -kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and ob- serve that the crate just begins to move when your force ex- ceeds 313 \(\mathrm{N}\) . After that you must reduce your push to 208 \(\mathrm{N}\) to keep it moving at a steady 25.0 \(\mathrm{cm} / \mathrm{s}\) . (a) What are the coefficients of static and kinetic friction between the crate and the floor? (b) What push must you exert to give it an acceleration of 1.10 \(\mathrm{m} / \mathrm{s}^{2}\) (c) Suppose you were performing the same experiment on this crate but were doing it on the moon instead, where the acceleration due to gravity is 1.62 \(\mathrm{m} / \mathrm{s}^{2} .\) (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

the task of designing an accelerometer to be used inside a rocket ship in outer space. Your equipment consists of a very light spring that is \(15.0 \mathrm{~cm}\) long when no forces act to stretch or compress it, \(\begin{array}{lllll}\text { plus } & \text { a } & 1.10 & \text { kg } & \text { weight. }\end{array}\) be attached to a friction- free tabletop, while the \(1.10 \mathrm{~kg}\) weight is attached to the other end, as shown in Figure 5.67 . (Such a spring-type accelerometer system was actually used in the ill-fated Genesis Mission, which collected particles of the solar wind. Unfortunately, because it was installed backward, it did not measure the acceleration correctly during the craft's descent to earth. As a result, the parachute failed to open and the capsule crashed on Sept. \(8,2004 .)\) (a) What should be the force constant of the spring so that it will stretch by \(1.10 \mathrm{~cm}\) when the rocket accelerates forward at \(2.50 \mathrm{~m} / \mathrm{s}^{2}\) ? Start with a free-body diagram of the weight. (b) What is the acceleration (magnitude and direction) of the rocket if the spring is compressed by \(2.30 \mathrm{~cm} ?\)

\(\bullet$$\bullet\) You push with a horl- zontal fonce of 50 \(\mathrm{N}\) against a 20 \(\mathrm{N}\) box, press- ing it against a rough ver- tical wall to hold it in place, The coefficients of kinetic and static friction between this box and the wall are 0.20 and 0.50 , respectively. (a) Make a free-body diagram of this box. (b) What is the friction force on the box? (c) How hard would you have to press for the box to slide downward with a uni- form speed of 10.5 \(\mathrm{cm} / \mathrm{s}\) ?

\(\bullet$$\bullet\) You are working for a shipping company. Your job is to stand at the bottom of an 8.0 -m-long ramp that is inclined at \(37^{\circ}\) above the horizontal. You grab packages off a conveyor belt and propel them up the ramp. The coefficient of kinetic friction between the packages and the ramp is \(\mu_{k}=0.30\) (a) What speed do you need to give a package at the bottom of the ramp so that it has zero speed at the top of the ramp? (b) Your coworker is supposed to grab the packages as they arrive at the top of the ramp, but she misses one and it slides back down. What is its speed when it returns to you?
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