Question

In loading a delivery van a driver pushed a 15 kg crate for 5 metres up a ramp. He had to push with a force of 132N and the vertical height gained was 1.3m. What was the efficiency of pushing this crate up the ramp onto the back of the van

Answer 1

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

$F\times d$ $=132\times5.17=682.44 joules$

$efficiency=\dfrac{ 195}{682.44}=0.29$

$efficiency$ %= $0.29\times100= 29$ %