Q24E - University Physics with Modern Physics

Question

You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground , it is travelling at 25.0 m/s. Use the work-energy theorem to find (a) the rock’s speed just as it left the ground and (b) its maximum height.

Step-by-Step Solution

Answer

(a) 30.3 m/s  

 

(b) 46.8 m  

Step 1: Identification of the given data

The given data is listed below as-

• The height of the rock above the ground is $h=15\hspace{0.33em}m$ 
• The final speed of rock is, $V_2=25\hspace{0.33em}m/s$ 

Step 2: Significance of the work-energy theorem

When forces act on a particle it undergoes displacement and the particle’s kinetic energy changes by an amount equal to the total work done on the particle by all the forces. Therefore, work done on the particle is given by-

${\mathbf{W}}_{\mathbf{total}}\mathbf{=}{\mathbf{K}}_{\mathbf{2}}\mathbf{-}{\mathbf{K}}_{\mathbf{1}}\mathbf{=}\mathbf{∆}\mathbf{K}$ 

The work-energy theorem can be applied to all the bodies that can be treated as particles.

Step 3: Determination of rock speed

(a)

Use the work-energy theorem,

$$\begin{gathered} W_{net}=K_2-K_1 \\ Fh=\frac{1}{2}m{V}_2^2-\frac{1}{2}m{V}_1^2 \end{gathered}$$ 

  

Now, the work done is negative since gravitational force is opposite to the direction of displacement.

$-mgh=\frac{1}{2}m{V}_2^2-\frac{1}{2}m{V}_1^2$ 

Divide the above equation by m to obtain the equation as below:

$-gh=\frac{1}{2}{V}_2^2-\frac{1}{2}{V}_1^2$ 

Here, m is the gravitational constant, h is height of the rock above ground, V₁ is the initial velocity of rock, and V₂ is the final velocity of the rock.

For, $g=9.8m{s}^{-2} , h=15 m , \mathrm{and} V_2=25m/s$

$-gh=\frac{1}{2}{V}_2^2-\frac{1}{2}{V}_1^2$ will be

$\left(9.8 {\mathrm{ms}}^{-2}\right)=\frac{1}{2}{\left(25 {\mathrm{ms}}^{-1}\right)}^2-\frac{1}{2}{\mathrm{V}}_1^2$ 

Therefore, solve the above equation and obtain the value of V₁ as below:

$V_1=30.0 m/s$  

 

Thus, the speed of rock as it just left the ground is 30.3m/s. 

Step 4: Determination of maximum height

(b)

The velocity at maximum height will be 0.

$-mg{h}_{max}=\frac{1}{2}m{V}_2^2-\frac{1}{2}m{V}_1^2$ 

Divide the above equation by m to obtain the equation as below:

 $-g{h}_{max}=\frac{1}{2}m{V}_2^2-\frac{1}{2}m{V}_1^2$

Here, m is the gravitational constant, V₁ is the initial velocity of rock, and V₂ is the final velocity of the rock.

For, $g=9.8m{s}^{-2} , V_2=0 , \mathrm{and} V_2=25m/s$ 

$-g{h}_{max}=\frac{1}{2}{V}_2^2-\frac{1}{2}{V}_1^2$  will be

$$\begin{gathered} -\left(9.8 m{s}^{-2}\right)h_{max}=0-\frac{1}{2}{\left(30.3 m{s}^{-1}\right)}^2 \\ {h}_{max}=46.8 m \end{gathered}$$ 

  

Therefore, the maximum height is 46.8 m