Your physics teacher drives an ideal frictionless car of mass 1000 kg which coasts along the level road to his home at a constant speed of 48 km/h with the engine turned off. It then coasts 40 m up his sloping driveway and comes to rest at the top without any braking required. One day he found that his engine had to develop 30 kilowatts of power to maintain a steady speed of 48 km/h on the level road due to a layer of loose snow on the road. He quickly calculated the new speed required to for him to coast up his driveway with the engine switched off as usual. He assumed that the snow on the hill would give the same constant retarding force as the snow on the level. What was the new speed?
To find the new speed required to coast up the driveway with the engine switched off, we need to consider the work-energy principle.
1. First, let's calculate the power developed by the engine to maintain a steady speed of 48 km/h on the level road. The power developed by the engine is given as 30 kilowatts.
Power (P) = 30 kilowatts = 30,000 watts
Speed (v) = 48 km/h = 48,000 meters/3,600 seconds = 13.33 m/s
2. Recall that power is defined as the rate at which work is done.
Power (P) = Work (W) / Time (t)
3. The work done by the engine is equal to the force applied by the engine multiplied by the distance traveled.
Work (W) = Force (F) × Distance (d)
4. Rearranging the equation for power, we can find the force applied by the engine on the level road.
Force (F) = Power (P) / Speed (v)
Force (F) = 30,000 watts / 13.33 m/s ≈ 2,250 N
5. Next, let's analyze the scenario of coasting up the sloping driveway with the engine switched off. We assume that the retarding force due to loose snow on the hill is equal to the retarding force due to snow on the level road.
6. The net force acting on the car on the hill is the difference between the gravitational force pulling it downhill and the retarding force due to the snow on the hill.
Net Force (F_net) = m × a
Since the car comes to rest at the top of the hill, the net force is zero. Therefore,
(m × a) = 0
7. The gravitational force pulling the car downhill is given by the equation:
Gravitational Force (F_gravity) = m × g × sin(θ)
Where m is the mass of the car (1000 kg), g is the acceleration due to gravity (9.8 m/s²), and θ is the angle of the driveway (which we do not know yet).
8. The retarding force due to the snow on the hill is equal to the force applied by the engine on the level road (2,250 N).
Retarding Force (F_retardation) = Force (F) = 2,250 N
9. Equating the gravitational force and the retarding force, we get:
m × g × sin(θ) = 2,250 N
10. Rearranging the equation, we can solve for sin(θ):
sin(θ) = (2,250 N) / (m × g)
sin(θ) = (2,250 N) / (1000 kg × 9.8 m/s²)
sin(θ) ≈ 0.23
11. Finally, to find the angle θ, we can use the inverse sine function:
θ ≈ sin^(-1) (0.23)
θ ≈ 13.2°
12. Since the car is coasting up the hill without any braking, the net work done on the car by the retarding force is zero. Therefore, the net work due to the retarding force is given by:
Net Work (W_net) = Retarding Force (F_retardation) × Distance (d)
13. We can calculate the distance d using the gravitational force equation:
Distance (d) = (W_net) / (F_gravity) = (0) / (m × g × sin(θ))
Distance (d) = 0
14. The new speed required to coast up the driveway with the engine switched off is the speed at which the car reaches the top of the hill, which is 0 m/s.
Therefore, the new speed is 0 m/s.