Part (1)
The free body daigram of the block on the inclined plane can be shown as,
According to work energy theorem,
[tex]W=\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
The work done on the block can be expressed as,
[tex]W=Td-mg\sin \theta d[/tex]
Plug in the known expression,
[tex]\begin{gathered} Td-mg\sin \theta d=\frac{1}{2}m(0)^2-\frac{1}{2}mu^2 \\ d(T-mg\sin \theta)=-\frac{mu^2}{2} \\ d=-\frac{mu^2}{2(T-mg\sin \theta)} \end{gathered}[/tex]
Substitute the known values,
[tex]\begin{gathered} d=-\frac{(16kg)(5.293m/s)^2}{2((50N)(\frac{1kgm/s^2}{1\text{ N}})-(16kg)(9.8m/s^2)\sin 39} \\ =-\frac{448.25kgm^2s^{-2}}{2(50kgm/s^2-98.7kgm/s^2)} \\ =-\frac{448.25kgm^2s^{-2}}{2(-48.7kgm/s^2)} \\ \approx4.60\text{ m} \end{gathered}[/tex]
Thus, the distance travelled by the block is 4.60 m.
Part (2)
The work done by gravitational force is given as,
[tex]W=-mg\sin \theta d[/tex]
Substitute the known values,
[tex]\begin{gathered} W=-(16kg)(9.8m/s^2)\sin 39(4.60\text{ m)(}\frac{1\text{ J}}{1kgm^2s^{-2}}) \\ =-(721.28\text{ J)(}0.629) \\ =-453.7\text{ J} \end{gathered}[/tex]
Thus, the work done by gravitational force is -453.7 J.
