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. 25 %
d. 360,000 %
c. 25 %
25% is the correct answer
To find the efficiency of the ramp, we need to divide the output work by the input work and then multiply by 100.
Efficiency = (Output work / Input work) x 100
Given that the input work is 1200 joules and the output work is 300 joules,
Efficiency = (300 / 1200) x 100
= 0.25 x 100
= 25 %
Therefore, the efficiency of the ramp is 25%, which is option c.
To calculate the efficiency of the ramp, we can use the formula:
Efficiency = (output work / input work) * 100
First, let's find the output work. The question states that only 300 joules of work is output for moving the box.
Output work = 300 joules
Next, let's find the input work. The question states that Tito and Michael put 1200 joules of work into pushing the box up the ramp.
Input work = 1200 joules
Now, let's substitute these values into the efficiency formula:
Efficiency = (300 joules / 1200 joules) * 100
Efficiency = (0.25) * 100
Efficiency = 25%
Therefore, the efficiency of the ramp is 25%.
The answer is: c. 25%