A 0.42 kgshuffleboard disk is initially at rest when a player uses a cue to increase its speed to 4.2 m/sat constant acceleration. The acceleration takes place over a 2.0 mdistance, at the end of which the cue loses contact with the disk. Then the disk slides an additional 12 mbefore stopping. Assume that the shuffleboard court is level and that the force of friction on the disk is constant. What is the increase in the thermal energy of the disk–court system (a) for that additional 12 mand (b) for the entire 14 mdistance? (c) How much work is done on the disk by the cue?

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:

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

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:

$$\begin{gathered} E_{th}=0.31\times 14 \\ =4.34J \\ \approx 4.3J \end{gathered}$$

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 .