Answer this questio

Dear student,
Refer below for above question:-
$$\begin{gathered} Tan A =\frac{2}{25}=0.08 \\ A = Ta{n}^{-1} \left(0.08\right) \\ A=4.{57}^0 \\ {F}_P= m.g. SinA \\ {F}_P=1000\times Sin 4.57 \\ {F}_P=797.5 N \\ Here F_P refers force parallel to incline. \\ Given V=18 Km/hr \\ V=18\times \frac{5}{18} \\ V=5 m/s \\ P=F_P\times V \\ P=797.5\times 5 \\ P=3988 J/s \\ Hence Power is approximately 4 KW \end{gathered}$$

Option 1 is correct.


Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards