Physics Efficiency Question

Question

A family uses several planks to slide a 350 kg piano onto the back of a pickup truck. The box on the back of the truck is 81 cm above the ground and the planks are 3.0 m long. An average force of magnitude 1500 N is required to slide the piano up the planks. Calculate the efficiency of the planks as a simple machine to load the piano.

Answer 1

Efficiency questions deal with energy and the ratio of the total change in energy of a process versus the work put into the process.

The goal of this problem is to raise the piano against the force of gravity. Gravity is a conservative force, so to calculate the work needed requires the starting and ending height of the piano.

Work is calculated as a force through a distance, or W = FΔs. For gravity, the force F is equal to mg (m, the mass of the piano, is 350 kg. g, the acceleration of gravity is a constant 9.8 m/s²). The change in distance Δs through which gravity works is the change in height, 81 cm or .81m.

Multiplying these variables out, the result for output is W = FΔs = mgΔh = 350kg * 9.8 m/s² * .81m = 2778.3J

For the input work, the family pushes the piano across planks. In the work formula W = FΔs, the force is given as F = 1500N. This force is applied across the length of the planks, which are 3m long, so Δs = 3m. The total work is W = FΔs = 1500N * 3m = 4500J.

Calculating the ratio of output/input gives the efficiency, % = Out/In = 2778.3J/4500J = .617 = 61.7%

So for the 4500J of work the family used to push the piano, 61.7% of it was used to move it against the force of gravity. The rest was likely lost as heat.