length of the ramp ="\\sqrt{(5^2+(1.3)^2)}=\\sqrt{(25+1.69)}=5.17m"
efficiency= "\\dfrac{net \\ work \\ done }{total\\ work \\ done}"
net work done is equal to the change in the potential energy of the box is
"mgh" "=15\\times10\\times1.3=195 \/ joules"
total work done is equal to work done by the driver up the plank is
"F\\times d" "=132\\times5.17=682.44 joules"
"efficiency=\\dfrac{ 195}{682.44}=0.29"
"efficiency % =" %= "0.29\\times100= 29" %
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