Problem 31 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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): \(\sin(3.00^{\circ}) = \frac{h}{2500}\)

Step 2: 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: \(h \approx 130.9\mathrm{~m}\)

Step 3: 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 \(1.93 \cdot 10^6\mathrm{~J}\).

Step 4: 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 \(1.93 \cdot 10^6\mathrm{~J}\).