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Chapter 6: Problem 71

A motorist driving a 1200 -kg car on level ground accelerates from $20.0 \mathrm{m} / \mathrm{s}\( to \)30.0 \mathrm{m} / \mathrm{s}\( in a time of \)5.0 \mathrm{s}$ Neglecting friction and air resistance, determine the average mechanical power in watts the engine must supply during this time interval.

Short Answer

Expert verified
Answer: The average mechanical power the engine must supply during this time interval is 60000 W.

Step by step solution

01

Calculate the acceleration of the car

To calculate the acceleration of the car, we will use the formula: Acceleration \((a) = \frac{\Delta v}{\Delta t}\) where \(\Delta v\) is the change in velocity and \(\Delta t\) is the time taken. The change in velocity is given by the final velocity \((v_f)\) minus the initial velocity \((v_i)\). \(\Delta v = v_f - v_i = 30.0 \mathrm{m/s} - 20.0 \mathrm{m/s} = 10.0 \mathrm{m/s}\) Using the formula, we get: \(a = \frac{\Delta v}{\Delta t} = \frac{10.0 \mathrm{m/s}}{5.0 \mathrm{s}} = 2.0 \mathrm{m/s^{2}}\)
02

Determine the force required for this acceleration

Now, we'll use Newton's second law of motion to find the force \((F)\) required for the car to have this acceleration. Newton's second law states that \(F = ma\), where \(m\) is the mass of the car and \(a\) is its acceleration. Given, mass of the car \((m) = 1200 \mathrm{kg}\) and acceleration \((a) = 2.0 \mathrm{m/s^{2}}\) \(F = (1200 \mathrm{kg})(2.0 \mathrm{m/s^{2}}) = 2400 \mathrm{N}\)
03

Calculate the distance traveled during this time interval

To calculate the distance traveled during this time interval, we'll use the formula: Average Velocity \((\bar{v}) = \frac{v_i + v_f}{2}\) \(\bar{v} = \frac{20.0 \mathrm{m/s} + 30.0 \mathrm{m/s}}{2} = 25.0 \mathrm{m/s}\) Now, we'll use this average velocity to compute the distance covered during this time interval using the formula: Distance \((d) = \bar{v} \times \Delta t\) \(d = (25.0 \mathrm{m/s})(5.0 \mathrm{s}) = 125 \mathrm{m}\)
04

Determine the work done using the force and the distance

The work done \((W)\) is the force multiplied times the distance the force has been applied. So, \(W = F \times d = (2400 \mathrm{N})(125 \mathrm{m}) = 300000 \mathrm{J}\)
05

Calculate the mechanical power using the work done and the time taken

Finally, we'll use the formula for power \((P)\): \(P = \frac{W}{\Delta t} = \frac{300000 \mathrm{J}}{5.0 \mathrm{s}} = 60000 \mathrm{W}\) The average mechanical power the engine must supply during this time interval is \(60000 \mathrm{W}\).

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Most popular questions from this chapter

A \(1500-\mathrm{kg}\) car coasts in neutral down a \(2.0^{\circ}\) hill. The car attains a terminal speed of \(20.0 \mathrm{m} / \mathrm{s} .\) (a) How much power must the engine deliver to drive the car on a level road at $20.0 \mathrm{m} / \mathrm{s} ?$ (b) If the maximum useful power that can be delivered by the engine is \(40.0 \mathrm{kW},\) what is the steepest hill the car can climb at \(20.0 \mathrm{m} / \mathrm{s} ?\)
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