Q27E - University Physics with Modern Physics

Question

A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest on a horizontal floor. It is then pushed in a straight line for by a trained dog that exerts a horizontal force with magnitude 36.0 N. Use the work–energy theorem to find the final speed of the 12-pack if 

(a) there is no friction between the 12-pack and the floor, and 

(b) the coefficient of kinetic friction between the 12-pack and the floor is 0.30.

Step-by-Step Solution

Answer

a) The final velocity of the pack without friction is 4.48 m/s.

b) The final velocity of the pack when friction is present is 3.61 m/s.

Step 1: Identification of given data

The given data can be listed below,

• The number of colas is, N = 12.
• The force on the cola is, F = 36 N.
• The distance travelled by dog is, s = 1.20 m.

Step 2: Concept/Significance of frictional force

The frictional force is defined as the force operating in the opposite direction between two surfaces. It is not a force of conservatism.

Step 3: Determination of the final speed of the 12-pack if (a) there is no friction between the 12-pack and the floor

The free body diagram of the system is given by,

The work done without frictional force is given by,

$$\begin{gathered} W_{tot}=\overrightarrow{F}.\overrightarrow{s} \\ =Fs \cos \varphi \end{gathered}$$ 

Here, F is the force on the cord, and s is the displacement.

 

The total work done on the system is given by,

 

$$\begin{gathered} W_{tot}=K_2-K_1 \\ =\frac{1}{2}m{v}_2^2 \end{gathered}$$

 

On comparing two equations, the final velocity of the pack is given by,

 

$$\begin{gathered} Fs=\frac{1}{2}m{v}_2^2 \\ {v}_2=\sqrt{\frac{2Fs}{m}} \end{gathered}$$ 

 

Substitute all the values in the above,

 

$$" width="9" height="19" style="max-width: none; vertical-align: -4px;" >$$\begin{gathered} v_2=\sqrt{\frac{Fs}{m}} \\ =\sqrt{\frac{2\times 36 N\left(1.20 m\right)}{4.30 kg}} \\ =\sqrt{20.09{\left(m/s\right)}^2} \\ =4.48 m/s \end{gathered}$$

 

Thus, the final velocity of the pack without friction is 4.48 m/s.

Step 4: (b) Determination of the final speed of the 12-pack if the coefficient of kinetic friction between the 12-pack and the floor is 0.30

The work done on the pack when the kinetic friction is acting on the system is given by,

 

$$\begin{gathered} W_{tot}=Fs-f_{k}s \\ =Fs-{\mu }_{k}mgs \end{gathered}$$

 

Also, the total work done is given by,

$$\begin{gathered} W_{tot}=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2 \\ =\frac{1}{2}m{v}_2^2-0 \end{gathered}$$ 

Comparing both the equations, the velocity of the pack when friction is applied is given by,

 

$$\begin{gathered} Fs-{\mu }_k mgs=\frac{1}{2} m{v}_2^2 \\ {v}_2^2=\frac{2(Fs-{\mu }_{k}mgs)}{m} \\ {v}_2 \\ =\sqrt{\frac{2(Fs-\mu \_k mgs)}{m}} \end{gathered}$$

 

Here, F is the force acting on the pack, ${\mu }_k$ is the coefficient of kinetic friction, and m is the mass of the pack.

 

Substitute all the values in the above,

 

$$\begin{gathered} v_2=\sqrt{\frac{2\left(Fs-{\mu }_{k}mgs\right)}{m}} \\ =\sqrt{\frac{2\left(36 N\times 1.2 m-0.30 kg\left(9.8 m/s^2\right)1.2 m\right)}{4.3 kg}} \\ =\sqrt{13.03}m/s \\ =3.61 m/s \end{gathered}$$ 

 

Thus, the final velocity of the pack when friction is present is 3.61 m/s.