Question

A 0.42 kg shuffleboard disk is initially at rest when a player uses a cue to increase its speed to 4.2 m/s at constant acceleration. The acceleration takes place over a 2.0 m distance, at the end of which the cue loses contact with the disk. Then the disk slides an additional 12 m before 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 m and (b) for the entire 14 m distance? (c) How much work is done on the disk by the cue?

Answer 1

Answer

given,

mass of the shuffleboard disk = 0.42 kg

speed of the cue is increased to = 4.2 m/s

acceleration takes over 2 m then acceleration is zero.

the disk additionally slide to 12 m

final speed of disk = 0 m/s

a) increase in thermal energy

$\Delta E_t = \dfrac{1}{2}m(v_1^2-v_2^2)$

$\Delta E_t = \dfrac{1}{2}\times 0.42 \times (4.2^2-v_2^2)$

$\Delta E_t = 3.704\ J$

b)  $\Delta E_t = F_f.d$

F_f is the frictional force

     $3.704= 12.F$

      $F_f = 0.308\ N$

increase in thermal energy for entire movement of 14 m

     $\Delta E_t = 0.308\times 14$

     $\Delta E_t =4.312 \ J$

c)  Work done on the disk by the cue

  $W = \Delta KE + \Delta E_{t}$

  $W = \dfrac{1}{2}\times 0.42 \times (4.2^2-0^2)+ F_f \times d$

  $W = 3.704+ 0.308 \times 2$

   W = 3.704 + 0.616

   W  = 4.32 J