Q57E - University Physics with Modern Physics

Question

Your job is to lift 30-kg crates a vertical distance of 0.90 m from the ground onto the bed of a truck. How many crates would you have to load onto the truck in 1 minute (a) for the average power output you use to lift the crates to equal 0.50 hp; (b) for an average power output of 100 W?

Step-by-Step Solution

Answer

1. The number of crates loaded with an average power of 0.50 hp is $84.58 {\min }^{-1}$ 
2. The number of crates loaded with an average power of 100 W is $22.68 {\min }^{-1}$ .

Step 1: Identification of given data

The given data can be listed below,

• The mass of the crate is, $m=30 kg$ 
• The distance of the crate from the ground on the bed of the truck is $d=0.90 m$  

Step 2: Concept/Significance of average power

Average power is calculated by dividing the total energy utilised by the total time required.

Step 3: (a) Determination of the number of crates loaded in 1 minute for the average power output of 0.50 hp

The number of crates loaded in one minute is given by,

$$\begin{gathered} N_{1min}=\frac{W_{1min}}{mgd} \\ =\frac{P_{av}}{mgd} \end{gathered}$$ 

Here, m is the mass of the crate, $W_{1min}$ is the total work done in one minute, g is the acceleration due to gravity, and d is the distance at, the crate is moved.

 

Substitute all the values in the above, 

$$\begin{gathered} N_{1min}=\frac{\left(0.50 \mathrm{hp}\right)\left(746 \mathrm{W}/\mathrm{hp}\right)}{30 \mathrm{kg}\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(0.9 \mathrm{m}\right)} \\ =1.41 {\mathrm{s}}^{-1}\left(\frac{60 \mathrm{s}}{1 \min }\right) \\ =84.58 {\min }^{-1} \end{gathered}$$ 

Thus, the number of crate loaded in 1 minute for average power 0.50 hp is $84.58 {\min }^{-1}$ .

Step 4: (b) Determination of number of crates loaded in 1 minute for the average power output 100 W

The number of crates loaded in one minute is given by,

$$\begin{gathered} N_{1min}=\frac{W_{1min}}{mgd} \\ =\frac{P_{av}}{mgd} \end{gathered}$$ 

Substitute all the values in the above,

$$\begin{gathered} N_{1min}=\frac{\left(100 \mathrm{W}\right)}{30 \mathrm{kg}\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(0.9 \mathrm{m}\right)} \\ =0.38 {\mathrm{s}}^{-1} \left(\frac{60 \mathrm{s}}{1 \min }\right) \\ =22.68 {\min }^{-1} \end{gathered}$$  

Thus, the number of crates loaded in 1 minute for average power 100 W  is  $22.68 {\min }^{-1}$