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A loaded truck of mass \(3000 \mathrm{~kg}\) moves on a level road at a constant speed of \(6.000 \mathrm{~m} / \mathrm{s}\). The frictional force on the truck from the road is \(1000 \mathrm{~N}\). Assume that air drag is negligible. (a) How much work is done by the truck engine in \(10.00 \mathrm{~min}\) ? (b) After \(10.00 \mathrm{~min}\), the truck enters a hilly region whose inclination is \(30^{\circ}\) and continues to move with the same speed for another \(10.00 \mathrm{~min}\). What is the total work done by the engine during that period against the gravitational force and the frictional force? (c) What is the total work done by the engine in the full \(20 \mathrm{~min}\) ?

Short Answer

Expert verified
(a) 3,600,000 J, (b) 56,520,000 J, (c) 60,120,000 J

Step by step solution

01

Understand the problem

We need to find the work done by the truck engine in two scenarios: (a) moving on a level road for 10 minutes at a constant speed, and (b) moving up a hill with a 30-degree incline for another 10 minutes at the same speed.
02

Calculate the work done in level road travel

Since the truck moves at a constant speed on a level road, the work done by the engine only overcomes friction. Use the formula: \[W = F \times d\]where \(F = 1000 \, \text{N}\) is the frictional force and \(d = v \times t\) is the distance traveled. Given the speed \(v = 6.000 \, \text{m/s}\) and time \(t = 10 \times 60 \, \text{s} = 600 \, \text{s}\), we calculate:\[\text{Distance} = 6.000 \, \text{m/s} \times 600 \, \text{s} = 3600 \, \text{m} \]Thus, the work done is:\[\text{Work} = 1000 \, \text{N} \times 3600 \, \text{m} = 3,600,000 \, \text{J (Joules)}\]
03

Calculate the work done moving uphill

In the hilly region, the engine does work against both friction and gravity. First, find the work against gravity:1. Calculate the vertical height ascended: \[\text{Height} = \text{distance} \times \sin(30^\circ) = 3600 \, \text{m} \times 0.5 = 1800 \, \text{m}\]2. Work against gravity is: \( \text{Gravitational force} = m \times g = 3000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \) \[W_g = 3000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1800 \, \text{m} = 52,920,000 \, \text{J}\]Next, calculate work against friction during uphill travel in the same manner as in Step 2:\[\text{Work against friction} = 1000 \, \text{N} \times 3600 \, \text{m} = 3,600,000 \, \text{J}\]Total work done uphill:\[W = W_f + W_g = 3,600,000 \, \text{J} + 52,920,000 \, \text{J} = 56,520,000 \, \text{J}\]
04

Calculate the total work done

To find the total work done in the full 20 minutes, sum up the work done in both segments:\[\text{Total work} = 3,600,000 \, \text{J} + 56,520,000 \, \text{J} = 60,120,000 \, \text{J}\]
05

Verification and Conclusion

Double-check each calculation step for accuracy and ensure consistent unit usage throughout the solution. The complete solution provides the answers to each sub-question in terms of Joules, the standard unit of work.

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Key Concepts

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

Frictional Force
Frictional force is a force that opposes the motion of objects sliding against each other. It acts in the opposite direction to the movement of the object. In the case of the loaded truck, as it travels on a level road, it has to continuously overcome this opposing force to maintain a constant speed.
This frictional force is exerted by the road on the truck, and it is given as 1000 N. Even though this may seem like a large force, the truck's engine is powerful enough to counteract it, allowing the truck to travel smoothly.
Understanding that the engine has to constantly do work not just to start the truck moving but also to keep it moving at a constant speed helps to grasp the notion of how frictional forces consume energy. This energy consumption is visible by the work done to maintain motion, which in this scenario, is calculated as the product of the frictional force and the distance traveled.
Gravitational Force
The gravitational force acts on objects with mass, pulling them towards the center of the Earth. In our scenario, gravity plays a significant role when the truck travels uphill after moving on a level road. Gravity's influence increases as the incline becomes steeper, requiring more work to climb than on flat terrain.
When a vehicle like the truck moves up an incline, its engine must do additional work against this gravitational pull. The gravitational force the truck faces is computed as the product of its mass and the acceleration due to gravity, which is approximately 9.8 m/s².
The work done against gravity further involves calculating the vertical distance ascended. The truck’s movement up the inclined plane is resisted by gravity, hence the need for more energy use to maintain the same speed uphill. As the vertical height entered was 1800 m on an incline, an understanding of how gravitational work is calculated deepens the awareness of energy expenditure during such uphill travels.
Constant Speed
The concept of constant speed means that the velocity of an object remains unchanged over time - there are no accelerations or decelerations. For the truck, constant speed implies that the forces propelling and retarding its motion (like friction and gravity, respectively) are balanced.
Achieving and maintaining constant speed are crucial in this scenario, as it simplifies calculations related to work done. Since speed is steady, the truck's kinetic energy doesn't increase or decrease, meaning that all the work done by the engine is used to counteract external forces, primarily friction on a flat road and gravitational force uphill.
Understanding constant speed also highlights how balanced forces come into play. The truck maintains its speed even on an incline because the work done exactly matches the work required to overcome both friction and gravity, demonstrating the requirement for equilibrium in these scenarios.
Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It eases the effort needed to lift an object by spreading the work over a longer distance, though in this scenario, the inclined plane represents a hill the truck must climb.
When traversing an inclined plane, an additional component of gravitational force acts parallel to the surface, resisting the upward movement. This requires the truck's engine to do extra work to counteract gravity, apart from overcoming friction. The incline's angle ( 30° in this case) determines the steepness and thus the difficulty of the climb. The steeper the incline, the more work needs to be done to ascend at constant speed.
In practice, understanding how inclined planes function is critical for calculating work done in real-world scenarios, like the one provided. It helps us recognize how slopes transform forces and necessitate additional energy, which the engine must supply for sustained movement.

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Most popular questions from this chapter

A single conservative force \(F(x)\) acts on a \(1.0 \mathrm{~kg}\) particle that moves along an \(x\) axis. The potential energy \(U(x)\) associated with \(F(x)\) is given by $$ U(x)=-4 x e^{-x / 4} \mathrm{~J} $$ where \(x\) is in meters. At \(x=5.0 \mathrm{~m}\) the particle has a kinetic energy of \(2.0 \mathrm{~J}\). (a) What is the mechanical energy of the system? (b) Make a plot of \(U(x)\) as a function of \(x\) for \(0 \leq x \leq 10 \mathrm{~m}\), and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine (c) the least value of \(x\) the particle can reach and (d) the greatest value of \(x\) the particle can reach. Use part (b) to determine (e) the maximum kinetic energy of the particle and (f) the value of \(x\) at which it occurs. (g) Determine an expression in newtons and meters for \(F(x)\) as a function of \(x\). (h) For what (finite) value of \(x\) does \(F(x)=0\) ?

Figure 8-23 shows a thin rod, of length \(L=2.00 \mathrm{~m}\) and negligible mass, that can pivot about one end to rotate in a vertical circle. A ball of mass \(m=5.00 \mathrm{~kg}\) is attached to the other end. The rod is pulled aside to angle \(\theta_{0}=30.0^{\circ}\) and released with initial velocity \(\vec{v}_{0}=0\). As the ball descends to its lowest point, (a) how much work does the gravitational force do on it and (b) what is the change in the gravitational potential energy of the ball-Earth system? (c) If the gravitational potential energy is taken to be zero at the lowest point, what is its value just as the ball is released? (d) Do the magnitudes of the answers to (a) through (c) increase, decrease, or remain the same if angle \(\theta_{0}\) is increased?

A conservative force \(\vec{F}=(6.0 x-12) \hat{i} \mathrm{~N}\), where \(x\) is in meters, acts on a particle moving along an \(x\) axis. The potential energy \(U\) associated with this force is assigned a value of \(27 \mathrm{~J}\) at \(x=0\). (a) Write an expression for \(U\) as a function of \(x\), with \(U\) in joules and \(x\) in meters. (b) What is the maximum positive potential energy? At what (c) negative value and (d) positive value of \(x\) is the potential energy equal to zero?

At \(t=0\) a \(1.0 \mathrm{~kg}\) ball is thrown from a tall tower with \(\vec{v}=(18 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(24 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). What is \(\Delta U\) of the ball-Earth system between \(t=0\) and \(t=6.0 \mathrm{~s}\) (still free fall)?

You drop a \(2.00 \mathrm{~kg}\) book to a friend who stands on the ground at distance \(D=10.0 \mathrm{~m}\) below. Your friend's outstretchedhands are at distance \(d=1.50 \mathrm{~m}\) above the ground (Fig. 8-19). (a) How much work \(W_{g}\) does the gravitational force do on the book as it drops to her hands? (b) What is the change \(\Delta U\) in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy \(U\) of that system is taken to be zero at ground level, what is \(U\) (c) when the book is released, and (d) when it reaches her hands? Now take \(U\) to be \(100 \mathrm{~J}\) at ground level and again find (e) \(W_{g}\), (f) \(\Delta U\), (g) \(U\) at the release point, and (h) \(U\) at her hands.

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