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Chapter 5: Problem 46

An engine expends 40.0 hp in moving a car along a level track at a speed of \(15.0 \mathrm{~m} / \mathrm{s}\). How large is the total force acting on the car in the opposite direction of the motion of the car?

Short Answer

Expert verified
Answer: The total force acting on the car in the opposite direction of its motion is approximately 1,988.54 N.

Step by step solution

01

Convert horsepower to watts

First, we need to convert the power from horsepower (hp) to watts (W). Use the conversion factor 1 hp = 745.7 W. Multiply 40.0 hp by 745.7 W/hp to get the power in watts. \(40.0 \, \text{hp} \times 745.7 \, \frac{\text{W}}{\text{hp}} = 29,828 \, \text{W}\)
02

Calculate force

Now that the power is in watts, we'll use the formula for power to find the force acting in the opposite direction of the motion: Power = Force × Speed We know the power (29,828 W) and the speed (15.0 m/s). Therefore, we can rewrite the formula to solve for force: Force = Power / Speed Plug in the values: Force = \(\frac{29,828 \, \text{W}}{15.0 \, \text{m/s}} = 1,988.54 \, \text{N}\)
03

Present the final result

The total force acting on the car in the opposite direction of its motion is approximately 1,988.54 N.

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Most popular questions from this chapter

A \(65-\mathrm{kg}\) hiker climbs to the second base camp on Nanga Parbat in Pakistan, at an altitude of \(3900 \mathrm{~m}\), starting from the first base camp at \(2200 \mathrm{~m}\). The climb is made in \(5.0 \mathrm{~h}\). Calculate (a) the work done against gravity, (b) the average power output, and (c) the rate of energy input required, assuming the energy conversion efficiency of the human body is \(15 \%\)

A certain tractor is capable of pulling with a steady force of \(14 \mathrm{kN}\) while moving at a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). How much power in kilowatts and in horsepower is the tractor delivering under these conditions?

Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars increase their speed by \(5.0 \mathrm{~m} / \mathrm{s}\), they then have the same kinetic energy, Calculate the original speeds of the two cars.

A particle moves parallel to the \(x\) -axis. The net force on the particle increases with \(x\) according to the formula \(F_{x}\) \(=(120 \mathrm{~N} / \mathrm{m}) x,\) where the force is in newtons when \(x\) is in meters. How much work does this force do on the particle as it moves from \(x=0\) to \(x=0.50 \mathrm{~m} ?\) a) 7.5 J c) \(30 J\) e) 120 J b) 15 J d) 60

Which of the following is a correct unit of energy? a) \(\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}\) c) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}^{2}\) e) \(\mathrm{kg}^{2} \mathrm{~m}^{2} / \mathrm{s}^{2}\) b) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}\) d) \(\mathrm{kg}^{2} \mathrm{~m} / \mathrm{s}^{2}\)
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