Answer
a) $t=5.3s$
b) increase
c) will stay the same
d) $F_t=mv_{\circ}$
Work Step by Step
(a) We know that
$a=-\mu_K g$
$a=-(0.13)(9.8m/s^2)$
$a=-1.274m/s^2$
Now $v=v_{\circ}+at$
$\implies 0m/s=(6.7m/s)+(-1.274m/s^2)t$
This simplifies to:
$t=5.3s$
(b) We know that the frictional force on the puck is directly proportional to the mass of the puck. Thus, if the mass of the puck is doubled then the frictional force on the puck is also doubled.
(c) We know that the stopping time depends on the coefficient of kinetic friction between the ice and the puck, initial speed of the puck and the acceleration due to gravity and doesn't depend on the mass of the puck. Therefore, there is no change in stopping time when the mass of the puck is doubled.
(d) We know that
$F_t=\mu_k mgt$
We plug in the known values to obtain:
$F_t=(0.13)(0.12Kg)(9.8m/s^2)(5.259s)$
$F_t=0.804Ns$
As $mv_{\circ}=(0.12Kg)(6.7m/s)=0.804Ns$ thus $F_t=mv_{\circ}$
