Tito and Michael put 1200 joules of work into pushing a box up a ramp, but only 300 joules of work is output for moving the box. What is the efficiency of the ramp?
a
4 %
b
25 %
c
360,000 %
d
.25 %
25% is the correct answer
To find the efficiency of the ramp, we need to use the formula:
Efficiency = (Output Work / Input Work) x 100
Given:
Input Work = 1200 joules
Output Work = 300 joules
Using the formula, we can calculate the efficiency:
Efficiency = (300 / 1200) x 100
Efficiency = (0.25) x 100
Efficiency = 25%
So, the efficiency of the ramp is 25%. Therefore, the correct answer is option b) 25%.
To find the efficiency of the ramp, we need to use the formula for efficiency:
Efficiency = (Output work / Input work) * 100%
In this case, the output work is 300 joules and the input work is 1200 joules. Plugging these values into the formula, we get:
Efficiency = (300 / 1200) * 100%
Simplifying the expression inside the parentheses:
Efficiency = 0.25 * 100%
Multiplying the decimal by 100%:
Efficiency = 25%
Therefore, the efficiency of the ramp is 25%.
Option b is the correct answer.