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Chapter 5: Problem 18

Think about the answers to these questions next time you are driving a car: a) What is the kinetic energy of a \(1500 .-\mathrm{kg}\) car moving at \(15.0 \mathrm{~m} / \mathrm{s} ?\) b) If the car changed its speed to \(30.0 \mathrm{~m} / \mathrm{s}\), how would the value of its kinetic energy change?

Short Answer

Expert verified
Answer: The kinetic energy increases by a factor of 4.

Step by step solution

01

Find the kinetic energy of the car moving at 15.0 m/s

To find the kinetic energy, we use the formula: $$ KE = \frac{1}{2}mv^2 $$ where m = 1500 kg and v = 15.0 m/s. By substituting these values, we can calculate the kinetic energy: $$ KE = \frac{1}{2}(1500 kg)(15.0 m/s)^2 $$
02

Calculate the value of KE1

Now, we calculate the value of kinetic energy: $$ KE = \frac{1}{2}(1500 kg)(225 m^2/s^2) $$ $$ KE = 168750 J $$ So, the kinetic energy of a car moving at 15.0 m/s is 168750 Joules.
03

Find the kinetic energy of the car moving at 30.0 m/s

Using the same formula, we will calculate the kinetic energy of the car when it changes its speed to 30.0 m/s: $$ KE = \frac{1}{2}mv^2 $$ where m = 1500 kg and v = 30.0 m/s. By substituting these values, we can calculate the kinetic energy: $$ KE = \frac{1}{2}(1500 kg)(30.0 m/s)^2 $$
04

Calculate the value of KE2

Now, we calculate the value of kinetic energy: $$ KE = \frac{1}{2}(1500 kg)(900 m^2/s^2) $$ $$ KE = 675000 J $$ So, the kinetic energy of a car moving at 30.0 m/s is 675000 Joules.
05

Compare the change in the kinetic energy values

We've found that the kinetic energy of the car when it's moving at 15.0 m/s is 168750 J and when it's moving at 30.0 m/s is 675000 J. To find the change, we can simply compare these two values: $$ \frac{KE2}{KE1} = \frac{675000 J}{168750 J} = 4 $$ So, when the car changes its speed from 15.0 m/s to 30.0 m/s, the value of its kinetic energy increases by a factor of 4.

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Most popular questions from this chapter

A \(95-\mathrm{kg}\) refrigerator rests on the floor. How much work is required to move it at constant speed for \(4.0 \mathrm{~m}\) along the floor against a friction force of \(180 \mathrm{~N} ?\)

A flatbed truck is loaded with a stack of sacks of cement whose combined mass is \(1143.5 \mathrm{~kg}\). The coefficient of static friction between the bed of the truck and the bottom sack in the stack is \(0.372,\) and the sacks are not tied down but held in place by the force of friction between the bed and the bottom sack. The truck accelerates uniformly from rest to \(56.6 \mathrm{mph}\) in \(22.9 \mathrm{~s}\). The stack of sacks is \(1 \mathrm{~m}\) from the end of the truck bed. Does the stack slide on the truck bed? The coefficient of kinetic friction between the bottom sack and the truck bed is \(0.257 .\) What is the work done on the stack by the force of friction between the stack and the bed of the truck?

An arrow of mass \(m=88 \mathrm{~g}(0.088 \mathrm{~kg})\) is fired from a bow. The bowstring exerts an average force of \(F=110 \mathrm{~N}\) on the arrow over a distance \(d=78 \mathrm{~cm}(0.78 \mathrm{~m})\) Calculate the speed of the arrow as it leaves the bow.

A car does the work \(W_{\text {car }}=7.0 \cdot 10^{4} \mathrm{~J}\) in traveling a distance \(x=2.8 \mathrm{~km}\) at constant speed. Calculate the average force \(F\) (from all sources) acting on the car in this process.

A car of mass 942.4 kg accelerates from rest with a constant power output of 140.5 hp. Neglecting air resistance, what is the speed of the car after 4.55 s?
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