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Chapter 6: Problem 31

A \(1.50 \cdot 10^{3}-\mathrm{kg}\) car travels \(2.50 \mathrm{~km}\) up an incline at constant velocity. The incline has an angle of \(3.00^{\circ}\) with respect to the horizontal. What is the change in the car's potential energy? What is the net work done on the car?

Short Answer

Expert verified
Answer: The net work done on the car is approximately \(1.93 \cdot 10^6\mathrm{~J}\).

Step by step solution

01

Calculate Vertical Height

To find the vertical height, we can use the sine function in trigonometry. The sine function is defined as the ratio of the side opposite a given angle in a right triangle to the hypotenuse. In this case, the angle is \(3.00^{\circ}\) and the incline distance traveled by the car is \(2.50 \mathrm{~km}\). First, we need to convert the incline distance to meters: \(2.50 \mathrm{~km} = 2500 \mathrm{~m}\). Now, we can use the sine function to find the vertical height (h): \(\sin(3.00^{\circ}) = \frac{h}{2500}\)
02

Solve for Vertical Height

To find the vertical height (h), we multiply both sides of the equation by \(2500\): \(h = 2500 \sin(3.00^{\circ})\) Calculating the value of h: \(h \approx 130.9\mathrm{~m}\)
03

Calculate Change in Potential Energy

Now that we have the vertical height, we can calculate the change in potential energy using the formula: \(ΔPE = mgh\) Where: \(ΔPE\) is the change in potential energy \(m\) is the mass of the car (\(1.50 \cdot 10^{3}~ \mathrm{kg}\)) \(g\) is the acceleration due to gravity (\(9.81~\mathrm{m/s^2}\)) \(h\) is the vertical height (\(130.9\mathrm{~m}\)) Plugging in the values: \(ΔPE = (1.50 \cdot 10^{3}\mathrm{~kg})(9.81\mathrm{~m/s^2})(130.9\mathrm{~m}) \approx 1.93 \cdot 10^6\mathrm{~J}\) So, the change in the car's potential energy is approximately \(1.93 \cdot 10^6\mathrm{~J}\).
04

Determine the Net Work Done on the Car

Since the car travels up the incline at a constant velocity, the net work done on the car is equal to the change in potential energy: \(W_{net} = ΔPE = 1.93 \cdot 10^6\mathrm{~J}\) The net work done on the car is approximately \(1.93 \cdot 10^6\mathrm{~J}\).

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Most popular questions from this chapter

How much mechanical energy is lost to friction if a 55.0-kg skier slides down a ski slope at constant speed of \(14.4 \mathrm{~m} / \mathrm{s}\) ? The slope is \(123.5 \mathrm{~m}\) long and makes an angle of \(14.7^{\circ}\) with respect to the horizontal.

a) If you are at the top of a toboggan run that is \(40.0 \mathrm{~m}\) high, how fast will you be going at the bottom, provided you can ignore friction between the sled and the track? b) Does the steepness of the run affect how fast you will be going at the bottom? c) If you do not ignore the small friction force, does the steepness of the track affect the value of the speed at the bottom?

A uniform chain of total mass \(m\) is laid out straight on a frictionless table and held stationary so that one-third of its length, \(L=1.00 \mathrm{~m},\) is hanging vertically over the edge of the table. The chain is then released. Determine the speed of the chain at the instant when only one-third of its length remains on the table.

A roller coaster is moving at \(2.00 \mathrm{~m} / \mathrm{s}\) at the top of the first hill \((h=40.0 \mathrm{~m}) .\) Ignoring friction and air resistance, how fast will the roller coaster be moving at the top of a subsequent hill, which is \(15.0 \mathrm{~m}\) high?

A ball is thrown up in the air, reaching a height of \(5.00 \mathrm{~m}\). Using energy conservation considerations, determine its initial speed.
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