a 600 n box is pushed up a ramp that is 2m high and 5 m long, the person is pushing the box at a force of 300 n,what is the efficiency of a ramp ?

The gain in potential energy is 600*2 = 1200 J, and this is the minimum required work that would be required in a frictionless situation. The work that actually has to be done is F*X = 300*5 = 1500 J. The efficiency is therefore 1200/1500 = 80%

This doesn't look like a 6th grade question to me.

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To find the efficiency of a ramp, we first need to understand the definition of efficiency. Efficiency is the ratio of useful output to the total input. In this case, the useful output is the amount of work done on the box by the person pushing it, and the total input is the work required to lift the box vertically.

First, let's calculate the work done by the person pushing the box. Work is defined as force applied multiplied by the distance moved in the direction of the force. In this case, the force applied is 300 N, and the distance moved is the length of the ramp, which is 5 m. Thus, the work done by the person pushing the box is:

Work = Force × Distance = 300 N × 5 m = 1500 N·m

Next, let's calculate the work required to lift the box vertically. The vertical distance it is lifted is 2 m, and the weight of the box is given as 600 N. The work required to lift the box is:

Work = Force × Distance = 600 N × 2 m = 1200 N·m

Now, we can calculate the efficiency of the ramp. Efficiency is defined as the ratio of useful work output to the total work input, multiplied by 100 to express it as a percentage. So the efficiency is:

Efficiency = (Useful work output / Total work input) × 100
Efficiency = (1500 N·m / 1200 N·m) × 100
Efficiency ≈ 125%

Therefore, the efficiency of the ramp is approximately 125%.

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