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Chapter 6: Problem 88

A car, starting from rest, accelerates in the \(+x\) direction as in the figure. It has a mass of \(1.10 \times 10^{3} \mathrm{kg}\) and maintains an acceleration of \(+4.60 \mathrm{m} / \mathrm{s}^{2}\) for \(5.00 \mathrm{s}\). Assume that a single horizontal force (not shown) accelerates the vehicle. Determine the average power generated by this force.

Short Answer

Expert verified
The average power generated is 58.19 kW.

Step by step solution

01

Calculate the Final Velocity

To find the final velocity of the car, use the formula for velocity as a function of acceleration and time:\[ v = u + at \]Here, the initial velocity \( u = 0 \ \text{m/s} \) (starting from rest), acceleration \( a = 4.60 \ \text{m/s}^2 \), and time \( t = 5.00 \ \text{s} \). Substitute the values:\[ v = 0 + (4.60 \ \text{m/s}^2)(5.00 \ \text{s}) = 23.0 \ \text{m/s} \] Thus, the final velocity \( v \) is \( 23.0 \ \text{m/s} \).
02

Calculate Work Done

The work done by the force is equal to the change in kinetic energy. The kinetic energy \( KE \) of the car is given by:\[ KE = \frac{1}{2}mv^2 \]Substitute the mass \( m = 1.10 \times 10^3 \ \text{kg} \) and final velocity \( v = 23.0 \ \text{m/s} \):\[ KE = \frac{1}{2}(1.10 \times 10^3 \ \text{kg})(23.0 \ \text{m/s})^2 = \frac{1}{2}(1.10 \times 10^3)(529) \]Calculating further:\[ KE = \frac{1}{2}(581900) = 290950 \ \text{Joules} \] The work done is \( 290950 \ \text{J} \).
03

Calculate Average Power

Average power is the work done per unit time. The formula for power \( P \) is:\[ P = \frac{W}{t} \]where \( W = 290950 \ \text{J} \) (work done) and \( t = 5.00 \ \text{s} \).Substitute the values:\[ P = \frac{290950 \ \text{J}}{5.00 \ \text{s}} = 58190 \ \text{W} \]Therefore, the average power generated by the force is \( 58190 \ \text{W} \), or \( 58.19 \ \text{kW} \).

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Key Concepts

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

Acceleration
Acceleration is a key concept in physics that describes how the velocity of an object changes over time. When a car accelerates, its speed increases in the direction of the applied force. In our exercise, the vehicle starts from rest and experiences an acceleration of \(+4.60 \, \text{m/s}^2\) for a duration of \(5.00\, \text{s}\).
  • Initial velocity: The car begins at \(0\, \text{m/s}\), as it starts from rest.
  • Acceleration factor: The acceleration given is constant at \(+4.60 \, \text{m/s}^2\).
  • Time period: Acceleration occurs for \(5.00\, \text{s}\).
To find the final velocity \(v\), we use the equation \(v = u + at\). Plugging in the initial velocity \(u\), acceleration \(a\), and time \(t\), we find that \(v = 23.0 \, \text{m/s}\). This velocity results from gradually increasing the car's speed under a constant acceleration over time.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It depends on both the mass of the object and its velocity. For the accelerating car in this exercise, we calculate kinetic energy using the formula:
\[KE = \frac{1}{2}mv^2\]Here, \(m\) is the mass of the car (\(1.10 \times 10^{3}\, \text{kg}\)) and \(v\) is the final velocity (\(23.0\, \text{m/s}\)).
  • Mass: The car's mass is a significant factor in calculating kinetic energy.
  • Velocity: Since velocity is squared in the formula, it greatly impacts the final energy value.
Substituting into the equation, the kinetic energy becomes \(290950 \, \text{J}\). This value reflects the work required to accelerate the vehicle to the final speed, indicating how much energy the car has due to its motion.
Work Done
Work done is the energy transferred to or from an object via the application of force along a displacement. In this exercise, the work done on the car is equivalent to the change in its kinetic energy. We compute the work done using the kinetic energy calculated previously:
  • Formula: The work done by the force causing acceleration equals the kinetic energy, \(290950 \, \text{J}\).
  • Time duration: This work is done over \(5.00\, \text{s}\), reflecting the constant effort applied by the force.
Understanding work done helps us see how force application leads to energy changes in the system. In terms of power, which is work done over time, the average power generated by the force is calculated to be \(58190 \, \text{W}\) or \(58.19 \, \text{kW}\). This indicates the rate at which energy is being transferred in achieving the vehicle's acceleration.

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Most popular questions from this chapter

A Makeshift Elevator. While exploring an elaborate tunnel system, you and your team get lost and find yourselves at the bottom of a \(450-\mathrm{m}\) vertical shaft. Suspended from a thick rope (near the floor) is a large rectangular bucket that looks like it had been used to transport tools and debris up and down the tunnel. Mounted on the floor near one of the walls is a gasoline engine \((3.5 \mathrm{hp})\) that turns a pulley and rope, and a sign that reads "Emergency Lift." It is clear that the engine is used to drive the bucket up the shaft. On the wall next to the engine is a sign indicating that a full tank of gas will last exactly 15 minutes when the engine is running at full power. You open the engine's gas tank and estimate that it is \(1 / 4\) full, and there are no other sources of gasoline. (a) Assuming zero friction, if you send your team's lightest member (who weighs \(125 \mathrm{lb}\) ), and the bucket weighs \(150 \mathrm{lb}\) when empty, how far up the shaft will the engine take her (and the bucket)? Will it get her out of the mine? (b) Assuming an effective collective friction (from the pulleys, etc.) of \(\mu_{\mathrm{eff}}=0.10\) (so that \(F_{\mathrm{f}}=\mu_{\mathrm{eff}} M g,\) where \(M\) is the total mass of the bucket plus team member), will the engine (with a \(1 / 4\) -full tank of gas) lift her to the top of the shaft?

A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of its engines is \(2.3 \times\) \(10^{5} \mathrm{N} .\) In being launched from rest it moves through a distance of \(87 \mathrm{m}\) and has a kinetic energy of \(4.5 \times 10^{7} \mathrm{J}\) at lift-off. What is the work done on the jet by the catapult?

"Rocket Man" has a propulsion unit strapped to his back. He starts from rest on the ground, fires the unit, and accelerates straight upward. At a height of \(16 \mathrm{m},\) his speed is \(5.0 \mathrm{m} / \mathrm{s} .\) His mass, including the propulsion unit, has the approximately constant value of 136 kg. Find the work done by the force generated by the propulsion unit.

Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the drive force is shut off, the snowmobile coasts to a halt. The snowmobile and its rider have a mass of 136 kg. Under the influence of a drive force of \(205 \mathrm{N}\), it is moving at a constant velocity whose magnitude is \(5.50 \mathrm{m} / \mathrm{s}\). The drive force is then shut off. Find (a) the distance in which the snowmobile coasts to a halt and (b) the time required to do so.

You are moving into an apartment and take the elevator to the 6 th floor. Suppose your weight is \(685 \mathrm{N}\) and that of your belongings is \(915 \mathrm{N}\). (a) Determine the work done by the elevator in lifting you and your belongings up to the 6 th floor \((15.2 \mathrm{m})\) at a constant velocity. (b) How much work does the elevator do on you alone (without belongings) on the downward trip, which is also made at a constant velocity?
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