91Ó°ÊÓ

a) Mass of the disk, m = 0.42kg

b) Initial speed of the disk,$v_i=0m/s$

c) Final speed of the disk,$v_f=4.2m/s$

d) Distance at which acceleration takes place,$d=2m$

e) Additional distance of slide,$x=12m$

f) Distance,$x=14m$

We use the concept of the work-energy principle. For the given distance, we can find an increase in thermal energy by work done. To find an increase in thermal energy for the entire distance, we find the frictional force, and then using it, we can find the thermal energy increased.

Formulae:

The net work done by the body,$W=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$ (i)

The work done due to applied force, $W=F.d$ (ii)

We know work done is the change in kinetic energy of the Shuffleboard. Here, increase in thermal energy is work done by frictional force. Thus, the increased thermal energy of the system for 12m is given by using the given data in equation (i) as:

$$\begin{gathered} W_f=\frac{1}{2}\left(0.42\right){\left(4.2\right)}^2-\frac{1}{2}\left(0.42\right){\left(0\right)}^2 \\ =3.7J \end{gathered}$$

Hence, the value of the increase in thermal energy is.

The friction force acting in the system is given using the above energy value in equation (ii) as:

$$\begin{gathered} f×12=3.7 \\ f=\frac{3.7}{12} \\ =0.31J \end{gathered}$$

For the entire$d_1=14m$ , we get the increase in thermal energy as work done by frictional force is given using equation (ii) as:

$$\begin{gathered} E_{th}=0.31×14 \\ =4.34J \\ ≈4.3J \end{gathered}$$

Hence, the value of the increased thermal energy is 4.3 J .

Work done by the cue is the increase in thermal energy and work done by frictional force is given as follows:

$$\begin{gathered} W_{cue}=3.7+\left(0.31\right)\left(2.0\right) \\ =4.3J \end{gathered}$$

Hence, the value of the work done is 4.3 J .