A block of mass 5   kg starting from rest pulled up on a smooth incline plane making an angle of 30…

Let's consider the following free-body diagram:

From the above diagram, it can be written that

$F-mg\sin 30°=ma ...\left(1\right)$

Substitute the values of the known parameters into equation (1) and solve to calculate the force applied on the block.

$\begin{array}{rcl}F-50 \sin 30°& =& 5\times 1\\ & \Rightarrow & F=\left(5+25\right) \mathrm{N}\\ & =& 30 \mathrm{N}\end{array}$

The velocity of the block $\left(v\right)$ at $t=10 \mathrm{s}$ can be written as

$\begin{array}{rcl}v& =& 0+1 \mathrm{m} {\mathrm{s}}^{-2}\times 10 \mathrm{s}\\ & =& 10 \mathrm{m} {\mathrm{s}}^{-1}\end{array}$

Hence, the required power is given by

$\begin{array}{rcl}P& =& Fv\\ & =& 30 \mathrm{N}\times 10 \mathrm{m} {\mathrm{s}}^{-1}\\ & =& 300 \mathrm{W}\end{array}$