Question

problem: a 500 n load is pushed up an incline with a length of 10 m and a height of 4 m. if the actual effort applied is 250 n, determine the efficiency of the inclined plane. you must show your work to receive full credit. type response here g: u: e: s: s:

Explanation:

Step1: Calculate the useful work

Useful work \(W_{u}\) is the work done in lifting the load vertically. The formula for work is \(W = F\times d\), where \(F\) is the force and \(d\) is the distance. Here, the force is the weight of the load \(F = 500\ N\) and the vertical distance \(d=4\ m\). So \(W_{u}=500\times4 = 2000\ J\).

Step2: Calculate the actual work input

The actual work input \(W_{i}\) is the work done by the applied force. The applied force \(F_{a}=250\ N\) and the distance along the incline \(d_{a}=10\ m\). So \(W_{i}=250\times 10=2500\ J\).

Step3: Calculate the efficiency

The efficiency \(\eta\) of a machine is given by the formula \(\eta=\frac{W_{u}}{W_{i}}\times 100\%\). Substituting the values of \(W_{u}\) and \(W_{i}\), we get \(\eta=\frac{2000}{2500}\times 100\% = 80\%\).

Answer:

80%