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A 1.50-kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s, and at point \(B\) it has slowed to 1.25 m/s. (a) How much work was done on the book between \(A\) and \(B\)? (b) If \(-\)0.750 J of work is done on the book from \(B\) to \(C\), how fast is it moving at point \(C\)? (c) How fast would it be moving at \(C\) if \(+\)0.750 J of work was done on it from \(B\) to \(C\)?

Short Answer

Expert verified
(a) -6.19 J, (b) 1.00 m/s, (c) 1.50 m/s.

Step by step solution

01

Understanding Kinetic Energy at Points A and B

First, calculate the kinetic energy of the book at point A using the formula \( KE = \frac{1}{2} m v^2 \). Given mass \( m = 1.50 \) kg and velocity \( v = 3.21 \) m/s, the kinetic energy at point A, \( KE_A = \frac{1}{2} \times 1.50 \times (3.21)^2 \). Similarly, calculate the kinetic energy at point B with velocity \( v = 1.25 \) m/s using the same formula: \( KE_B = \frac{1}{2} \times 1.50 \times (1.25)^2 \).
02

Calculate Work Done from A to B

The work done on the book between points A and B is the change in its kinetic energy. This can be found using the formula \( W_{AB} = KE_B - KE_A \). Substitute the kinetic energies computed in the previous step to find \( W_{AB} \).
03

Determine Velocity at Point C with -0.750 J Work

To find the velocity at point C when \(-0.750\) J of work is done, first determine the new kinetic energy at C. \( KE_C = KE_B + W_{BC} \) where \( W_{BC} = -0.750 \) J. Solve for velocity \( v \) using \( KE_C = \frac{1}{2} m v^2 \).
04

Determine Velocity at Point C with +0.750 J Work

If \( +0.750 \) J of work is done, similarly calculate \( KE_C = KE_B + 0.750 \) J. Solve for the new velocity at point C using the kinetic energy formula \( KE_C = \frac{1}{2} m v^2 \).
05

Solve for Each Scenario

1. Work done from A to B is calculated. 2. Velocity at C with -0.750 J work is determined. 3. Velocity at C with +0.750 J work is found. These results provide a comprehensive understanding of the book's kinetic energy changes through work applied at different segments.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It can be calculated using the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. For the book in our exercise, we first determine its kinetic energy at point A using its mass of 1.50 kg and velocity of 3.21 m/s. This calculation helps us understand how much energy the book has at the start of the journey along the rough surface. Similarly, the kinetic energy at point B is calculated with the updated velocity of 1.25 m/s. By comparing these two kinetic energy values, we can track how the motion and energy of the book change over time. This is the foundation for understanding the work done on the book as it moves.
Work Done on an Object
Work done on an object is a measure of energy transfer that occurs when a force acts on the object to move it over a distance. When considering how much work was done on the book between points A and B, it's crucial to understand that it's determined by the change in kinetic energy. The formula utilized is \( W_{AB} = KE_B - KE_A \), highlighting that work done is essentially the difference in the book's kinetic energy between these two points. In our scenario, the calculated work reveals the amount of energy lost to friction on the rough surface, which causes the book to slow down as it approaches point B.
Velocity Change
A change in velocity occurs when an object speeds up or slows down as a result of forces acting on it. In our exercise, between points A and B, the book experiences a reduction in velocity due to the negative work done by friction. Afterwards, the problem explores how additional work done affects velocity at point C. Negative work, meaning work done against the motion, further reduces the kinetic energy and thus the velocity. Conversely, positive work adds energy, allowing the book to speed up. By understanding these scenarios, we appreciate how work, whether positive or negative, directly influences an object's speed and direction.
Conservation of Energy
The conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another. In our problem, we see the transition of energy from kinetic energy to work against external forces like friction. As the book transitions from point A to C through B, the work-energy theorem is applied to reflect these energy changes. In practical terms, the energy initially present in the book's motion is partially transformed into work done against friction, causing a velocity change. The book's final speed at point C depends on this energy transformation, showcasing the broader theme of energy conservation in automotive motion.

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Most popular questions from this chapter

A block of ice with mass 2.00 kg slides 1.35 m down an inclined plane that slopes downward at an angle of 36.9\(^\circ\) below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.

While doing a chin-up, a man lifts his body 0.40 m. (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the \(total\) percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.) (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 J of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.

An airplane in flight is subject to an air resistance force proportional to the square of its speed v. But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward (\(\textbf{Fig. P6.94}\)). The upward force is the lift force that keeps the airplane aloft, and the backward force is called \(induced \, drag\). At flying speeds, induced drag is inversely proportional to \(v^2\), so the total air resistance force can be expressed by \(F_air = \alpha v^{2} + \beta /v{^2}\), where \(\alpha\) and \(\beta\) are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna 150, a small single-engine airplane, \(\alpha = 0.30 \, \mathrm{N} \cdot \mathrm{s^{2}/m^{2}}\) and \(\beta = 3.5 \times 10^5 \, \mathrm{N} \cdot \mathrm{m^2/s^2}\). In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed (in km/h) at which this airplane will have the maximum \(range\) (that is, travel the greatest distance) for a given quantity of fuel. (b) Calculate the speed (in km/h) for which the airplane will have the maximum \(endurance\)(that is, remain in the air the longest time).

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On a farm, you are pushing on a stubborn pig with a constant horizontal force with magnitude 30.0 N and direction 37.0\(^\circ\) counterclockwise from the +\(x\)-axis. How much work does this force do during a displacement of the pig that is (a) \(\overrightarrow{s} = (5.00 \, \mathrm{m})\hat{\imath}\); (b) \(\overrightarrow{s} = - (6.00 \, \mathrm{m})\hat{\jmath}\); (c) \(\overrightarrow{s} = - (2.00 \, \mathrm{m})\hat{\jmath} + (4.00 \, \mathrm{m}) \hat{\jmath}\)?

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