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Chapter 5: Problem 82

Losing Cargo. A \(12.0-\mathrm{kg}\) box rests on the flat floor of a truck. The coefficients of friction between the box and floor are \(\mu_{s}=0.19\) and \(\mu_{k}=0.15 .\) The truck stops at a stop sign and then starts to move with an acceleration of 2.20 \(\mathrm{m} / \mathrm{s}^{2} .\) If the box is 1.80 \(\mathrm{m}\) from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck? How far does the truck travel in this time?

Short Answer

Expert verified
The box falls off the truck after about 2.22 seconds, and the truck travels approximately 5.42 meters in that time.

Step by step solution

01

Find the Maximum Static Friction Force

The maximum static friction force can be calculated using the formula: \( f_{s, \, max} = \mu_s \times m \times g \). Here, \( m = 12.0 \, \text{kg} \) is the mass of the box, \( \mu_s = 0.19 \) is the static friction coefficient, and \( g = 9.8 \, \text{m/s}^{2} \) is the gravitational acceleration. Therefore, \( f_{s, \, max} = 0.19 \times 12.0 \times 9.8 = 22.344 \, \text{N} \).
02

Compare Frictional Forces Against Truck's Force

Next, calculate the force exerted by the truck's acceleration on the box: \( F = m \times a \), where \( a = 2.20 \, \text{m/s}^{2} \) is the truck's acceleration. Thus, \( F = 12.0 \times 2.20 = 26.4 \, \text{N} \). Since \(26.4 \, \text{N} > 22.344 \, \text{N} \), static friction is overcome, and the box will slide.
03

Calculate Kinetic Friction Force

Now, calculate the kinetic friction force since the box is sliding. Use \( f_k = \mu_k \times m \times g \) where \( \mu_k = 0.15 \). Therefore, \( f_k = 0.15 \times 12.0 \times 9.8 = 17.64 \, \text{N} \).
04

Determine Net Force on the Box

The net force acting on the box is the difference between the force due to acceleration and the kinetic friction force: \( F_{\text{net}} = F - f_k = 26.4 - 17.64 = 8.76 \, \text{N} \).
05

Find Box's Acceleration

Use the net force to find the box's acceleration using \( F_{\text{net}} = m \times a_{\text{box}} \). So, \( a_{\text{box}} = F_{\text{net}} / m = 8.76 / 12.0 = 0.73 \, \text{m/s}^2 \).
06

Calculate Time to Slide Off the Truck

The box will travel back 1.80 m to reach the edge of the truck. Use the equation \( s = \frac{1}{2} a_{\text{box}} \times t^2 \) to solve for \( t \): \( 1.80 = \frac{1}{2} \times 0.73 \times t^2 \). Solving this gives \( t = \sqrt{\frac{1.80 \times 2}{0.73}} \approx 2.22 \, \text{seconds} \).
07

Calculate Distance Traveled by the Truck

Finally, find how far the truck travels in the time it takes for the box to slide off. Use the formula \( s = \frac{1}{2} a_{\text{truck}} \times t^2 \). Plugging in the known values, \( s = \frac{1}{2} \times 2.20 \times (2.22)^2 \approx 5.42 \, \text{meters} \).

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Key Concepts

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

Static Friction
Static friction is the force that keeps an object at rest when it is subject to an external force. This force needs to be overcome for an object to start moving. Static friction is given by the formula:
  • \( f_{s, \, max} = \mu_s \times m \times g \)
Here, \( \mu_s \) is the coefficient of static friction, \( m \) is the mass in kilograms, and \( g \) is the gravitational acceleration \( (9.8 \text{ m/s}^2) \). This equation calculates the maximum force needed to overcome the static friction.
The purpose of static friction is essentially to "glue" the object to the surface until enough force is applied to make it slide. In the given exercise, the truck exerts a force that exceeds the maximum static friction force, causing the box to start sliding.
Kinetic Friction
Kinetic friction comes into play once an object has started moving. Unlike static friction, kinetic friction remains constant regardless of the speed of movement. For kinetic friction, the formula is:
  • \( f_k = \mu_k \times m \times g \)
In this formula, \( \mu_k \) is the coefficient of kinetic friction. This coefficient is generally lower than the static coefficient, which explains why it takes less force to keep an object in motion than to start its motion.
In the exercise, after the static friction is overcome, kinetic friction dictates the opposing force on the box. This force is used to find how the box accelerates once it starts moving. Here, it is shown that the force of kinetic friction is lower than the static friction, demonstrating the concept that less force is needed to keep the box moving.
Newton's Second Law
Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The law is often written as:
  • \( F = m \times a \)
Where \( F \) is the net force in newtons, \( m \) is the mass in kilograms, and \( a \) is the acceleration in meters per second squared.
In the context of the exercise, once the static friction is overcome, you apply this principle to find the box's acceleration. The net force \, acting on the box was calculated by subtracting the kinetic friction force from the force exerted by the truck's acceleration. This allows us to determine how fast the box accelerates backwards as the truck accelerates forward.

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Most popular questions from this chapter

A bowling ball weighing 71.2 \(\mathrm{N}(16.0 \mathrm{lb})\) is attached to the ceiling by a \(3.80-\mathrm{m}\) rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.20 \(\mathrm{m} / \mathrm{s} .\) (a) What is the acceleration of the bowling ball, in magnitude and direction, at this instant? (b) What is the tension in the rope at this instant?

You find it takes 200 \(\mathrm{N}\) of horizontal force to move an empty pickup truck along a level road at a speed of 2.4 \(\mathrm{m} / \mathrm{s}\) . You then load the pickup and pump up its tires so that its total weight increases by 42\(\%\) while the coefficient of rolling friction decreases by 19\(\%\) . Now what horizontal force will you need to move the pickup along the same road at the same speed? The speed is low enough that you can ignore air resistance.

A physics student playing with an air hockey table (a frictionless surface) finds that if she gives the puck a velocity of 3.80 \(\mathrm{m} / \mathrm{s}\) along the length \((1.75 \mathrm{m})\) of the table at one end, by the time it has reached the other end the puck has drifted 2.50 \(\mathrm{cm}\) to the right but still has a velocity component along the length of 3.80 \(\mathrm{m} / \mathrm{s} .\) She correctly concludes that the table is not level and correctly calculates its inclination from the given information. What is the angle of inclination?

You are working for a shipping company. Your job is to stand at the bottom of a 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 \(\mu_{\mathbf{X}}=0.30\) . (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?

A flat (unbanked) curve on a highway has a radius of 220.0 \(\mathrm{m}\) . A car rounds the curve at a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) , (a) What is the minimum coefficient of friction the coefficient of friction (b) Suppose the highway is icy and the coefficient of friction between the tires and pavement is only one-third what you found in part (a). What should be the maximum speed of the car so it can round the curve safely?
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