A wheel and axle is used to raise a load of 500 N by the application of an effort of 250 N. If the radii of the wheel and the axle are 0.4 cm and 0.1 cm res...

A wheel and axle is used to raise a load of 500 N by the application of an effort of 250 N. If the radii of the wheel and the axle are 0.4 cm and 0.1 cm respectively, the efficiency of the machine is

Answer Details

A wheel and axle is a simple machine consisting of two circular objects of different sizes. The wheel is the larger circular object and the axle is the smaller one that is connected to the wheel. The load is attached to the axle, and the effort is applied to the wheel to raise the load. The efficiency of a machine is defined as the ratio of output work to input work, expressed as a percentage. In this case, the output work is the work done on the load, which is equal to the product of the load lifted and the height it is lifted to. The input work is the work done by the effort, which is equal to the product of the effort applied and the distance moved by the effort. To calculate the output work, we need to determine the distance moved by the load when it is raised by one revolution of the wheel. This distance is equal to the circumference of the axle, which is given by 2πr, where r is the radius of the axle. Therefore, the distance moved by the load in one revolution of the wheel is 2π x 0.1 cm = 0.628 cm. To calculate the input work, we need to determine the distance moved by the effort when the wheel makes one revolution. This distance is equal to the circumference of the wheel, which is given by 2πr, where r is the radius of the wheel. Therefore, the distance moved by the effort in one revolution of the wheel is 2π x 0.4 cm = 2.512 cm. The load lifted is 500 N, and the effort applied is 250 N. The height to which the load is lifted is equal to the distance moved by the load in one revolution of the wheel, which is 0.628 cm. Therefore, the output work is 500 N x 0.628 cm = 314 J. The work done by the effort is equal to the product of the effort applied and the distance moved by the effort, which is 250 N x 2.512 cm = 628 J. The efficiency of the machine is the ratio of output work to input work, expressed as a percentage. Therefore, the efficiency of the machine is (314 J / 628 J) x 100% = 50%. Therefore, the closest answer option to the efficiency of the machine is 50%.