Question

A ramp is 3.0 meters long and 1.5 meters high. The distant of the effort, therefore, is the length of the ramp and the resistance distance is the height of the ramp. If it takes 60.0 N of force to move a crate that weighs 105 N up the ramp, what is the efficiency of the ramp?

Answer 1

88%

Step-by-step explanation:

Answer 2

For the given problem, the calculated efficiency of the ramp is 87.5 %.

Explanation:

The efficiency of any instrument or machine is defined as the percentage of ratio of work obtained from the instrument for a given amount of input work. So in this case, the efficiency of the ramp can be calculated as the ratio of real mechanical advantage to ideal mechanical advantage.

The ideal mechanical advantage is the measure of the work done by us on the crater. So it can be calculated as the ratio of distance covered for a given effort also known as effort distance to the resistance distance of the ramp.

And the real mechanical advantage is the measure of work done by the crater for the given input force. So, it is calculated as the ratio of resistance force to effort force.

For the given situation, the effort distance is the length of the ramp i.e. 3 m and the resistance distance is the height of the ramp i.e., 1.5 m. And the effort force is the amount of force required to perform a work i.e., 60 N and the resistance force is the normal force acting on the crate i.e., 105 N.

So, the mathematical representation of all the parameters are as follows:

`\begin{array}{c}{\text { Efficiency }=\frac{\text { Real Mechanical Advantage }}{\text { Ideal Mechanical Advantage }} * 100 \%} \\ {\text { Real Mechanical Advantage }=\frac{\text { Resistance force }}{\text { Effort force }}=(105)/(60)=1.75}\end{array}`

`\begin{array}{l}{\text { Ideal Mechanical Advantage }=\frac{\text { Effort distance }}{\text { Resistance distance }}=(3)/(1.5)=2} \\ {\text {Efficiency }=(1.75)/(2) * 100 \%=87.5 \%}\end{array}`

Thus, the efficiency of the ramp is 87.5 %.