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Chapter 7: Problem 30

While a roofer is working on a roof that slants at \(36^{\circ}\) above the horizontal, he accidentally nudges his \(85.0 \mathrm{~N}\) toolbox, causing it to start sliding downward from rest. If it starts \(4.25 \mathrm{~m}\) from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is \(22.0 \mathrm{~N}\) ?

Short Answer

Expert verified
The toolbox will be moving at a speed calculated from Step 3 as it reaches the edge of the roof.

Step by step solution

01

Resolve Forces into Components

First, resolve the weight of the toolbox into components parallel and perpendicular to the roof. The component of the toolbox weight along the slope of the roof gives the force causing the toolbox to slide and it can be computed as \(85.0 \mathrm{~N} \times \sin(36^\circ)\). The frictional force, which opposes the motion, has a magnitude of 22.0 N. Therefore, the net force acting on the toolbox is the difference between these two.
02

Calculate Acceleration

Next, apply Newton's second law to find the acceleration of the toolbox. According to the law, the net force on an object is equal to the product of its mass and acceleration. Therefore, acceleration can be obtained by dividing the net force by the mass of the toolbox. The weight of the toolbox is its mass times the acceleration due to gravity, so mass can be calculated as \(85.0 \mathrm{~N} / 9.81 \mathrm{~m/s^2}\). Hence, divide the net force obtained in Step 1 by this mass to find the acceleration.
03

Calculate Final Velocity

Lastly, use the kinematic equation to find the final velocity. The equation of motion that suits this situation is \(V_f^2 = V_i^2 + 2ad\), where \(V_f\) is final velocity, \(V_i\) is initial velocity, \(a\) is acceleration, and \(d\) is distance. The toolbox initially is at rest so \(V_i = 0\). Substitute the values of \(a\) from Step 2 and \(d = 4.25 \mathrm{~m}\) into the equation to solve for \(V_f\).

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Key Concepts

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

Kinematic Equations
Kinematic equations are the foundation of describing motion in physics. When objects move, whether across a table or down a slope, these equations help us predict their positions, velocities, and accelerations over time. For the roofer's sliding toolbox problem, one key kinematic equation is used: \( V_f^2 = V_i^2 + 2ad \), where \( V_f \) is the final velocity, \( V_i \) is the initial velocity, \( a \) is the acceleration, and \( d \) is the distance.

Since the toolbox starts from rest, the initial velocity \( V_i \) is zero. This simplifies our equation considerably. By plugging the acceleration \( a \) - found by applying Newton's second law - and the distance \( d = 4.25 \text{ m} \) that the toolbox slides, we can solve for the final velocity \( V_f \) to determine how fast it will be moving as it reaches the edge of the roof. Understanding and applying these kinematic equations correctly is vital for solving many problems in physics related to motion.
Newton's Second Law
Newton's second law is central to understanding motion and forces. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (\( F = ma \)). In our roofer's problem, the net force acting on the toolbox is what causes it to accelerate down the slope.

The total weight of the toolbox acts downwards due to gravity but only the component of this weight along the slope (\( 85.0 \, \text{N} \times \sin(36^\circ) \)) plays a role in sliding motion, and this force is opposed by the frictional force of \( 22.0 \, \text{N} \). The difference between these two gives us the net force. Subsequently, by dividing this net force by the mass of the toolbox (which we get from its weight divided by gravitational acceleration), we find the acceleration. Through a clear understanding of Newton's second law, students can analyze the motion of objects under various forces, enhancing their problem-solving abilities in physics.
Frictional Force
Frictional force is an omnipresent force that opposes the relative motion between two surfaces in contact. In the textbook problem, the rooftop acts as one surface and the bottom of the toolbox as the other. The magnitude of the kinetic frictional force is given as \(22.0 \, \text{N}\).

This force resists the motion induced by the gravitational component pulling the toolbox along the roof's slope. When calculating the acceleration of the toolbox in our problem, we must take into account the retarding effect of this frictional force. The net force that results after subtracting the frictional force from the downslope component of gravity is what actually contributes to the acceleration of the toolbox according to Newton's second law. Braking down complex concepts like friction into real-world problems such as this one helps students visualize and better comprehend the abstract principles of physics.

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Most popular questions from this chapter

\(\mathrm{A}\) spring of negligible mass has force constant \(k=800 \mathrm{~N} / \mathrm{m}\) (a) How far must the spring be compressed for \(1.20 \mathrm{~J}\) of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then lay a \(1.60 \mathrm{~kg}\) book on top of the spring and release the book from rest. Find the maximum distance the spring will be compressed.

BIO Human Energy vs. Insect Energy. For its size, the common flea is one of the most accomplished jumpers in the animal world. A 2.0 -mm-long, \(0.50 \mathrm{mg}\) flea can reach a height of \(20 \mathrm{~cm}\) in a single leap. (a) Ignoring air drag, what is the takeoff speed of such a flea? (b) Calculate the kinetic energy of this flea at takeoff and its kinetic energy per kilogram of mass. (c) If a \(65 \mathrm{~kg}, 2.0-\mathrm{m}\) -tall human could jump to the same height compared with his length as the flea jumps compared with its length, how high could the human jump, and what takeoff speed would the man need? (d) Most humans can jump no more than \(60 \mathrm{~cm}\) from a crouched start. What is the kinetic energy per kilogram of mass at takeoff for such a \(65 \mathrm{~kg}\) person? (e) Where does the flea store the energy that allows it to make sudden leaps?

\(\mathrm{A} 1500 \mathrm{~kg}\) rocket is to be launched with an initial upward speed of \(50.0 \mathrm{~m} / \mathrm{s} .\) In order to assist its engines, the engineers will start it from rest on a ramp that rises \(53^{\circ}\) above the horizontal (Fig. \(\mathbf{P 7 . 5 0}\) ). At the bottom, the ramp turns upward and launches the rocket vertically. The engines provide a constant forward thrust of \(2000 \mathrm{~N}\), and friction with the ramp surface is a constant \(500 \mathrm{~N}\). How far from the base of the ramp should the \text { rocket start, as measured along the surface of the ramp?

CALC The potential energy of two atoms in a diatomic molecule is approximated by \(U(r)=\left(a / r^{12}\right)-\left(b / r^{6}\right),\) where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) on one atom as a function of \(r .\) Draw two graphs: one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r\). (b) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. (d) For the molecule CO, the cquilibrium distance between the carbon and oxygen atoms is \(1.13 \times 10^{-10} \mathrm{~m}\) and the dissociation energy is \(1.54 \times 10^{-18} \mathrm{~J}\) per molecule. Find the values of the constants \(a\) and \(b\).

You are designing a delivery ramp for crates containing exercise equipment. The \(1470 \mathrm{~N}\) crates will move at \(1.8 \mathrm{~m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a \(515 \mathrm{~N}\) kinctic friction force on cach crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of \(5.0 \mathrm{~m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the largest force constant of the spring that will be needed to meet the design criteria.
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