Q50P - University Physics with Modern Physics

Question

You have landed on an unknown planet, Newtonian, and want to know what objects weigh there. When you push a certain tool, starting from rest, on a frictionless horizontal surface with a $\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{N}$ force, the tool moves $\mathbf{16}\mathbf{.0 }\mathbf{m}$ in the first $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{s}$. You next observe that if you release this tool from rest at $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{m}$ above the ground, it takes $\mathbf{2}\mathbf{.}\mathbf{58}\mathbf{ }\mathbf{s}$ to reach the ground. What does the tool weigh on Newtonian, and what does it weigh on Earth?

Step-by-Step Solution

Answer

The weight of the tool in Newtonian is $4.5 \mathrm{N}$.

The weight of the tool on Earth is $14.7 \mathrm{N}$.

Step 1: Identification of the given data:

The given data can be listed below as:

The force exerted on the tool is, $\mathrm{F}=12.0 \mathrm{N}$.

The distance moved by the tool initially is $\mathrm{s}=16.0 \mathrm{m}$.

The time taken by the tool to move initially is $\mathrm{t}=2.00 \mathrm{s}$.

The distance moved by the tool finally is ${\mathrm{s}}_1=10.0 \mathrm{m}$.

The time taken by the tool to move finally is ${\mathrm{t}}_1=2.58 \mathrm{s}$.

Step 2: Significance of the weight

The weight is described as the gravitational attraction force that helps an object attract another object. The weight is described as the product of the mass of that object and the acceleration due to gravity.

Step 3: Determination of the weight of the tool in Newtonia

The equation of the acceleration of the tool is expressed as:

 

 $$\begin{gathered} \mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{s}-\mathrm{ut}=\frac{1}{2}{\mathrm{at}}^2 \\ 2\left(\mathrm{s}-\mathrm{ut}\right)={\mathrm{at}}^2 \\ \mathrm{a}=\frac{2\left(\mathrm{s}-\mathrm{ut}\right)}{{\mathrm{t}}^2} \end{gathered}$$

                                                                                                            ….. (1)

Here, a is the acceleration of the tool, s is distance moved by the tool initially, u is the initial velocity of the tool and t is the time taken by the tool to move initially.

 

The equation of the mass of the tool is expressed as:

 

  $$\begin{gathered} \mathrm{F}=\mathrm{ma} \\ \mathrm{m}=\frac{\mathrm{F}}{\mathrm{a}} \end{gathered}$$
 
Here, F is the force exerted on the tool, m is the mass and a is the acceleration of the tool.

 

Substitute the values of the equation (1) in the above equation.

 

 $$\begin{gathered} \mathrm{m}=\frac{\mathrm{F}}{{\displaystyle \frac{2\left(\mathrm{s}-\mathrm{ut}\right)}{{\mathrm{t}}^2}}} \\ =\frac{{\mathrm{Ft}}^2}{2\left(\mathrm{s}-\mathrm{ut}\right)} \end{gathered}$$

 

As initially the tool was at rest, the initial velocity of the tool is zero.

 

Substitute all the values in the above equation.

 

 $$\begin{gathered} \mathrm{m}=\frac{\left(12.0\mathrm{N}\right){\left(2.00 \mathrm{s}\right)}^2}{2\left(16.0 \mathrm{m}-\left(0\right)\mathrm{t}\right)} \\ =\frac{\left(12.0\mathrm{N}\times \left({\displaystyle \frac{1 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1 \mathrm{N}}}\right)\right)\left(4.00 {\mathrm{s}}^2\right)}{\left(32.0 \mathrm{m}\right)} \\ =\left(12.0 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\right)\left(0.125 {\mathrm{s}}^2/\mathrm{m}\right) \\ =1.5 \mathrm{kg} \end{gathered}$$

 

The equation of the acceleration of the tool finally is expressed as:

 

 $$\begin{gathered} \mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{a}}_1{\mathrm{t}}^2 \\ \mathrm{s}-\mathrm{ut}=\frac{1}{2}{\mathrm{a}}_1{\mathrm{t}}^2 \\ 2\left(\mathrm{s}-\mathrm{ut}\right)={\mathrm{a}}_1{\mathrm{t}}^2 \\ {\mathrm{a}}_1=\frac{2\left(\mathrm{s}-\mathrm{ut}\right)}{{\mathrm{t}}^2} \end{gathered}$$

                                                                                                         ….. (2)

 

Here,${\mathrm{a}}_1$is the acceleration of the tool, s is distance moved by the tool finally, u is the initial velocity of the tool and t is the time taken by the tool to move finally.

 

The equation of the weight of the tool in Newtonian is expressed as:

 

  $\mathrm{W}={\mathrm{ma}}_1$

 

Here, W is the weight of the tool in Newtonian, m is the mass and${\mathrm{a}}_1$is the final acceleration of the tool.

 

Substitute the value from the equation (2) in the above equation.

 

 $\mathrm{W}=\mathrm{m}\frac{2\left(\mathrm{s}-\mathrm{ut}\right)}{{\mathrm{t}}^2}$

 

Substitute the values in the above equation.

 

 $$\begin{gathered} \mathrm{W}=\left(1.5 \mathrm{kg}\right)\frac{2\left(10.0 \mathrm{m}-\left(0\right)\mathrm{t}\right)}{\left(2.58 \mathrm{s}\right)} \\ =\left(1.5 \mathrm{kg}\right)\frac{\left(20.0 \mathrm{m}\right)}{\left(6.65 \mathrm{s}\right)} \\ =\left(1.5 \mathrm{kg}\right)\left(3.007 \mathrm{m}/{\mathrm{s}}^2\right) \\ =4.5 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2} \\ \mathrm{W}=4.5 \mathrm{N} \end{gathered}$$

 Thus, the weight of the tool in Newtonian is $4.5 \mathrm{N}$.

Step 4: Determination of the weight of the tool in Earth

The equation of the weight of the tool in Earth is expressed as:

 

 ${\mathrm{W}}_1=\mathrm{mg}$

 

Here,${\mathrm{W}}_1$is the weight of the tool on Earth, m is the mass and g is the acceleration due to gravity.

 

Substitute the values in the above equation.

 

 $$\begin{gathered} {\mathrm{W}}_1=\left(1.5 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ =14.7 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2 \\ =14.7 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right) \\ =14.7 \mathrm{N} \end{gathered}$$

 

Thus, the weight of the tool on Earth is  14.7 N.