/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 A car comes to a skidding stop i... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 10

A car comes to a skidding stop in \(15 \mathrm{~m}\). The force on the car due to the road is \(1000 \mathrm{~N}\). The work done by road on the car and car on the road, respectively, is (A) \(-15 \mathrm{~kJ}\), zero (B) zero, \(15 \mathrm{~kJ}\) (C) \(15 \mathrm{~kJ}\), zero (D) \(-15 \mathrm{~kJ}, 15 \mathrm{~kJ}\)

Short Answer

Expert verified
The work done by the road on the car is \(-15 \mathrm{~kJ}\), and the work done by the car on the road is \(15 \mathrm{~kJ}\). Therefore, the correct option is (D) \(-15 \mathrm{~kJ}, 15 \mathrm{~kJ}\).

Step by step solution

01

Identify the knowns and unknowns

We are given: - Distance the car comes to a stop: \(d = 15 \mathrm{~m}\) - Force on the car due to the road: \(F = 1000 \mathrm{~N}\) - We need to find the work done by the road on the car, and the work done by the car on the road.
02

Find the angle between force and distance

Since the car comes to a stop due to the force exerted by the road, the angle between the force and distance is 180掳 (opposite directions). Therefore, \(\theta = 180^\circ\).
03

Calculate the work done by the road on the car

Use the formula for work, \(W = Fd \cos{\theta}\), and plug in the given values: \(W_{road\_to\_car} = 1000(15)cos(180^\circ) = 1000(15)(-1) = -15000 \mathrm{Nm} \) Convert 15000 Nm to kJ: \(-15000\div1000 = -15 \mathrm{~kJ}\) Therefore, the work done by the road on the car is \(-15 \mathrm{~kJ}\).
04

Calculate the work done by the car on the road

According to Newton's Third Law of Motion, the force exerted by the car on the road is equal and opposite to the force exerted by the road on the car. Thus, the work done by the car on the road is equal and opposite to the work done by the road on the car: \(W_{car\_to\_road} = -(-15 \mathrm{kJ}) = 15 \mathrm{~kJ}\) Therefore, the work done by the car on the road is \(15 \mathrm{~kJ}\).
05

Determine the correct option

Based on our calculations, the work done by the road on the car is \(-15 \mathrm{~kJ}\) and the work done by the car on the road is \(15 \mathrm{~kJ}\). So, the correct option is: (D) \(-15 \mathrm{~kJ}, 15 \mathrm{~kJ}\)

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