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
[tex]\Delta E_t = \dfrac{1}{2}m(v_1^2-v_2^2)[/tex]
[tex]\Delta E_t = \dfrac{1}{2}\times 0.42 \times (4.2^2-v_2^2)[/tex]
[tex]\Delta E_t = 3.704\ J[/tex]
b) [tex]\Delta E_t = F_f.d[/tex]
F_f is the frictional force
[tex]3.704= 12.F[/tex]
[tex]F_f = 0.308\ N[/tex]
increase in thermal energy for entire movement of 14 m
[tex]\Delta E_t = 0.308\times 14[/tex]
[tex]\Delta E_t =4.312 \ J[/tex]
c) Work done on the disk by the cue
[tex]W = \Delta KE + \Delta E_{t}[/tex]
[tex]W = \dfrac{1}{2}\times 0.42 \times (4.2^2-0^2)+ F_f \times d[/tex]
[tex]W = 3.704+ 0.308 \times 2[/tex]
W = 3.704 + 0.616
W = 4.32 J
