Q55P - University Physics with Modern Physics

Question

You are a Starfleet captain going boldly where no man has gone before. You land on a distant planet and visit an engineering testing lab. In one experiment a short, light rope is attached to the top of a block and a constant upward force $\mathbf{ }\mathbf{F}$ is applied to the free end of the rope. The block has mass m and is initially at rest. As $\mathbf{F}$ is varied, the time for the block to move upward 
$\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}$ is measured. The values that you collected are given in the table:

                                       

(a) Plot $\mathbf{F}$ versus the acceleration   of the block. (b) Use your graph to determine the mass   of the block and the acceleration of gravity $\mathbf{g}$ at the surface of the planet. Note that even on that planet, measured values contain some experimental error.

Step-by-Step Solution

Answer

(a)                  

 

(b) The mass of the block is 25.6 kg.

The acceleration due to gravity at the surface of the planet is  $8.3 \mathrm{m}/{\mathrm{s}}^2$.

Step 1: Identification of the given data:

The given data can be listed below as,

• The force applied on the rope is $\mathrm{F}$.
• The mass of the block is $\mathrm{m}$.
• The distance moved by the block is $\mathrm{s}=8.00 \mathrm{m}$.

Step 2: Significance of the acceleration

The acceleration is described as the division of the force exerted on the object and the mass of that object. Moreover, the acceleration also helps to find the velocity of an object.

Step 3: (a) Determination of the force versus acceleration graph

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

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

Here, $\mathrm{a}$ is the acceleration of the block, $\mathrm{s}$ is the distance moved by the block, $\mathrm{u}$  is the initial speed of the block and $\mathrm{t}$ is the time taken by the block to reach to the desired distance.

 

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

Substitute $8.00 \mathrm{m}$ and  $\mathrm{s}$ and $0$ for $\mathrm{u}$ in the above equation.

 

 $$\begin{gathered} \mathrm{a}=\frac{2\left(\left(8.00 \mathrm{m}\right)-\left(0\right)\mathrm{t}\right)}{{\mathrm{t}}^2} \\ \mathrm{a}=\frac{16 \mathrm{m}}{{\mathrm{t}}^2} \end{gathered}$$

                                                                                                                    ….. (1)

In the first case,

Substitute  $3.3 \mathrm{s}$ for $\mathrm{t}$ in the above equation.

 $$\begin{gathered} \mathrm{a}=\frac{16 \mathrm{m}}{{\left(3.3 \mathrm{s}\right)}^2} \\ =\frac{16 \mathrm{m}}{\left(10.89 {\mathrm{s}}^2\right)} \\ =1.469 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

 

In the second case,

Substitute 2.2 s for t into equation (1).

 

 $$\begin{gathered} \mathrm{a}=\frac{16 \mathrm{m}}{{\left(2.2 \mathrm{s}\right)}^2} \\ =\frac{16 \mathrm{m}}{\left(4.84 {\mathrm{s}}^2\right)} \\ =3.306 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

 

In the third case,

Substitute $1.7 \mathrm{s}$ for $\mathrm{t}$ into equation (1).

 $$\begin{gathered} \mathrm{a}=\frac{16 \mathrm{m}}{{\left(1.7 \mathrm{s}\right)}^2} \\ =\frac{16 \mathrm{m}}{\left(2.89 {\mathrm{s}}^2\right)} \\ =5.536 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

 

In the fourth case,

Substitute $1.5 \mathrm{s}$ for t into equation (1).

 

 $$\begin{gathered} \mathrm{a}=\frac{16 \mathrm{m}}{{\left(1.5 \mathrm{s}\right)}^2} \\ =\frac{16 \mathrm{m}}{\left(2.25 {\mathrm{s}}^2\right)} \\ =7.111 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

 

In the fifth case,

 

Substitute $1.3 \mathrm{s}$ for $\mathrm{t}$ into equation (1).

 $$\begin{gathered} \mathrm{a}=\frac{16 \mathrm{m}}{{\left(1.3 \mathrm{s}\right)}^2} \\ =\frac{16 \mathrm{m}}{\left(1.69 {\mathrm{s}}^2\right)} \\ =9.467 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

 

In the sixth case,

Substitute  $1.2 \mathrm{s}$ for $\mathrm{t}$ into equation (1).

$$\begin{gathered} \mathrm{a}=\frac{16 \mathrm{m}}{{\left(1.2 \mathrm{s}\right)}^2} \\ =\frac{16 \mathrm{m}}{\left(1.44 {\mathrm{s}}^2\right)} \\ =11.11 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

From the above data, the graph of $\mathrm{F}$ versus $\mathrm{a}$ is expressed as:

 

               

Step 4: (b) Determination of the mass of the block:

The mass of the block can be obtained by observing the straight line of the graph. From the straight line, it can be observed that the starting line of the straight line starts from $212.98 \mathrm{N}$ as the acceleration starts from the point.

From the graph, the equation of the straight line is expressed as:

$\mathrm{F}=\left(25.6 \mathrm{kg}\right)\mathrm{a}+212.98 \mathrm{N}$                                                              ….. (2)

The equation of the force on the block is expressed as:

$\mathrm{F}=\mathrm{ma}+\mathrm{mg}$                                                                                      ….. (3)

Comparing the equation (2) and (3), the mass of the block can be expressed as:

 $\mathrm{m}=25.6 \mathrm{kg}$ 

Thus, the mass of the block is $25.6 \mathrm{kg}$.

Step 5: (b) Determination of the acceleration due to gravity of the block

Comparing the equation (2) and (3), the acceleration due to gravity of the block can be expressed as:

$\mathrm{mg}=212.98 \mathrm{N}$ 

Here, $\mathrm{m}$ is the mass of the block and $\mathrm{g}$ is the acceleration due to gravity of the block.

Substitute $25.6 \mathrm{kg}$ for $\mathrm{m}$ in the above equation.

$$\begin{gathered} \mathrm{g}=\frac{212.98 \mathrm{N}}{25.6 \mathrm{kg}} \\ =8.3 \mathrm{N}/\mathrm{kg} \\ =8.3 \mathrm{N}/\mathrm{kg}\times \frac{1 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}} \\ =8.3 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

Thus, the acceleration due to gravity at the surface of the planet is  $8.3 \mathrm{m}/{\mathrm{s}}^2$.