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Bob pushed a crate up a 3.0m long ramp to a truck bed with a force of 800n. The truck bed is 1.5m off the ground and the crate has a mass of 120kg calculate the following work output work input efficiency of the inclined plane

User Sekomer
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The output work can be calculated as the change in mechanical energy of the crate.

Since the height of the crate increased by 1.5 meters, the change in mechanical energy occurs by an increase in potential energy:


\begin{gathered} E_p=m\cdot g\cdot h \\ E_p=120\cdot10\cdot1.5 \\ E_p=1800\text{ J} \end{gathered}

The input work is given by the work done by Bob to move the crate, using a force of 800 N in a distance of 3 meters:


\begin{gathered} W=F\cdot d \\ W=800\cdot3 \\ W=2400\text{ J} \end{gathered}

The efficiency is given by the output energy over the input energy:


e=\frac{\text{output}}{\text{ input}}=(1800)/(2400)=0.75

So the efficiency is 75%.

User Rpgmaker
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