Question

Consider a 60.0 kg box that is pushed up an 8.0 m long ramp with a force of 100.0 N to a height of 1.0 m in 5 seconds. What is the efficiency % of the ramp?

Answer 1

$The\; formula \; of \; efficiency\; (\gamma_{percent} )\; =\;( W_{out}/W_{in})\times 100$

$W_{out}=mgh$

$W_{out}=60\times 9.8\times 1$

$\mathbf{W_{out}=588J}$

$\mathbf{Now\; we \; need \; to\; calculate W_{in}}$

$W_{in}}=Fd$

$W_{in}=Fd$

$W_{in}=100N\times 8m$

$\mathbf{Now \; use\; the\; formula\; of\; efficiency(\gamma_{percentage} )=(W_{out}/W_{in}})\times 100$

                                                             $efficiency(\gamma_{percentage} )=(588J/800J)\times 100$

                                                               $efficiency(\gamma_{percentage} )=(.735)\times 100$

                                                                $\mathbf{efficiency(\gamma_{percentage} )=(73.5)\; \; Percentage}$