A \(1500-\mathrm{kg}\) car coasts in neutral down a \(2.0^{\circ}\) hill. The car attains a terminal speed of \(20.0 \mathrm{m} / \mathrm{s} .\) (a) How much power must the engine deliver to drive the car on a level road at $20.0 \mathrm{m} / \mathrm{s} ?$ (b) If the maximum useful power that can be delivered by the engine is \(40.0 \mathrm{kW},\) what is the steepest hill the car can climb at \(20.0 \mathrm{m} / \mathrm{s} ?\)

#Analysis#: In this exercise, we need to find (a) the power the engine should deliver to drive the car at the given speed on a level road and (b) the steepest hill the car can climb at that speed, given the maximum useful power of the engine. For both parts, we will need to use equations related to work, power, and force, considering that the car's terminal velocity on a hill or a level road is affected by forces like gravity and air resistance. Solution: #tag_title# Part (a): Power needed to drive the car on a level road at 20 m/s #tag_content# The forces acting on a car driving on a level road at a constant speed are gravity, air resistance, and friction. The net force on the car will be zero since it is moving at a constant speed and not accelerating. However, the engine must deliver power equal to the opposing forces (air resistance and friction) to maintain that speed. 1. Calculate the gravitational force acting on the car: F_gravity = m * g where m = 1500 kg (mass of the car) and g = 9.81 m/s^2 (acceleration due to gravity) F_gravity = 1500 * 9.81 = 14,715 N 2. Calculate the force acting on the car due to the hill's inclination: F_inclination = m * g * sin(theta) where theta = 2.0 degrees (angle of the hill) F_inclination = 1500 * 9.81 * sin(2) = 517.4 N 3. Calculate the terminal velocity of the car: v_terminal = 20.0 m/s (given) 4. Calculate the air resistance (drag force) acting on the car: F_drag = 0.5 * C_d * A * rho * v_terminal^2 where C_d is the drag coefficient, A is the cross-sectional area of the car, rho is air density, and v_terminal is the terminal velocity. We will assume that the drag force is equal to the gravitational force acting parallel to the road. F_drag = 14,715 N 5. Calculate the power needed to drive at a constant speed: P_required = F_drag * v_terminal P_required = 14,715 * 20 = 294,300 W #tag_title# Part (b): Steepest hill the car can climb at 20 m/s #tag_content# To find the steepest hill the car can climb, given the maximum useful power of the engine, we can use the formula for power and the relationship between the gravitational force and the angle of inclination. 1. Convert the maximum useful power to watts: P_max = 40,000 W (given as 40 kW) 2. Calculate the maximum force that can be applied by the engine: F_max = P_max / v_terminal (where v_terminal = 20 m/s from Part (a)) F_max = 40,000 / 20 = 2,000 N 3. Calculate the maximum angle of inclination such that F_max = m * g * sin(theta_max): theta_max = arcsin (F_max / (m * g)) theta_max = arcsin (2,000 / (1500 * 9.81)) = 0.087 radians 4. Find the steepest hill the car can climb at 20 m/s: theta_max (in degrees) = 0.087 * (180 / pi) = 5.0 degrees