Answer
$(a)\space 195.6\space W$
$(b)\space 3320\space J$
$(c)\space 227.7\space W, 3864\space J$
Work Step by Step
Please see the attached image first.
(a) Here we use the equation $P=FV$ to find the power needed, where P- Power, F- Applied force, V- speed of the object.
First of all, let's find the applied force on the object.
$\rightarrow F=ma$ ; Let's plug known values into this equation.
$F-F_{f}= m(0)$
$F=F_{f}=\mu R=0.612\times 84.5g$
$F=506.8\space N$
$P=FV$ ; Let's plug known values into this equation.
$*P=506.8\space N\times0.386\space m/s=195.6\space W$
(b) Let's apply equation $W=FS$ to find the work done.
$*W= 506.8\space N\times 6.55\space m= 3320\space J$
(c) Please see figure (b) of the image
Let's apply equation $F=ma$ to the object.
$\angle5.75^{\circ} \nearrow F=ma$ ; Let's plug known values into this equation.
$F-F_{s}-84.8gsin5.75^{\circ}=0$
$F=\mu\times84.5\times g\times cos5.75^{\circ}+84.8\times g\times sin5.75^{\circ}$
$F=84.5\space kg\times9.8\space m/s^{2}(0.612\times1+0.1)$
$F=590\space N$
$P=FV$ ; Let's plug known values into this equation.
$*P=590\space N\times0.386\space m/s=227.7\space W$
Let's apply equation $W=FS$ to find the work done.
$*W=590\space N\times 6.55\space m= 3864\space J$
