What distance in the diagram is ##50\cos(8°)##m?physicsmaster123 said:
The work done by gravity on a car rolling down a hill is calculated as the product of the gravitational force acting on the car, the distance the car travels down the hill, and the cosine of the angle between the force and the direction of motion. Mathematically, it is expressed as W = mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the vertical height the car descends.
The angle of the hill affects the component of the gravitational force acting along the direction of the car's motion. A steeper hill means a larger component of the gravitational force is acting in the direction of the car's motion, resulting in more work done by gravity. However, the total work done by gravity depends only on the vertical height descended, not the angle of the hill.
Yes, the mass of the car directly influences the work done by gravity. Since the gravitational force is proportional to the mass of the car (F = mg), a heavier car will experience a greater gravitational force and thus more work will be done by gravity for the same vertical descent.
Friction does not directly affect the work done by gravity itself, but it does affect the net work done on the car. While gravity does work to increase the car's kinetic energy, friction does negative work, dissipating energy as heat and reducing the car's acceleration. The overall energy change of the car will be the work done by gravity minus the work done by friction.
No, the work done by gravity on a car rolling down a hill is always positive because the force of gravity and the direction of the car's motion are in the same direction. However, if the car were moving uphill against gravity, the work done by gravity would be negative, as gravity would be opposing the car's motion.