Question

In loading a delivery van, the driver pushed a 15 kg crate 5 meter up the ramp. He had to push with the force of 132 N and the vertical height gain was 1.3m. What was the efficiency $(\eta )$ of pushing this crate up the ramp onto the back of the van?

Answer 1

Hint: In order to solve this question, to find the efficiency of pushing this crate, first we will draw the diagram to see what question says and then we will apply the formula of efficiency for which we have to find the output work and the input work applied on the crate basically the input and the output work is the potential energies which are applied by gravitational force and the force we applied respectively.

Formula Used:
Efficiency of the crate can be calculated by this formula
\[\text{Efficiency}(\eta ) = \dfrac{\text{output work}}{\text{input work}} \times 100\% \]
Here, $(\eta )$ – refers to the efficiency (“eta” in Greek letter)
Output work – refers to potential energy which is produced in process.
Input work – refers to potential energy which we have to put in process.
$\text{output work} = \text{force} \times \text{vertical height}$
Here, Force – refers to how much force we applied on the crate and Vertical height – refers to what height the crate is when measured vertically.
$\text{Input work} = \text{force} \times \text{vertical height}$
Force – refers to gravitational force on crate
Vertical height – height of ramp vertically
${\text{potential energy = force }} \times {\text{ height}}$

Complete step by step answer:

The above diagram represents all the forces which are applying on the crate.First, we will find the output work produces. Output work is the potential energy we apply on the crate
Here, we are given that force applied on crate=132 N and Vertical height=1.3 m.
$\text{output work} = \text{force} \times \text{vertical height}$
$ \Rightarrow 132{\text{ N }} \times 1.3{\text{ m}}$
$ \Rightarrow 171.6{\text{ joules}}$
Now, we will find input work produces, input work is the potential energy which is applied by the gravitational force. Here, we are given that
${\text{force applied = mass of crate}} \times {\text{g}}$
$g$: gravitational constant (9.8)
vertical height of ramp = 5m

$\text{Input work} = \text{force} \times \text{vertical height}$
$ \Rightarrow \text{Input work} = 15{\text{ }}kg \times {\text{g}} \times {\text{5m}}$
$ \Rightarrow \text{Input work} = 15 \times 9.8 \times 5$
$ \Rightarrow \text{Input work} = 735\,joules$
The Efficiency $(\eta )$ of pushing this crate up the ramp onto the back of the van
\[\text{Efficiency}(\eta ) = \dfrac{\text{output work}}{\text{input work}} \times 100\% \]
$ \Rightarrow \text{Efficiency}(\eta ) =\dfrac{{171.6}}{{725}}$
$ \therefore \text{Efficiency}(\eta ) =23.34\% $

Therefore, the efficiency of pushing this crate up the ramp onto the back of the van is $23.34\% $.

Note: Efficiency of anything is basically how much energy we can conserve in the process. In other words we are comparing the input of the energy by the output of energy. For example, in the process energy is wasted in other forms which we don’t need for example heat and light this reduces the efficiency.