The work done in pulling up a block of wood weighing 2kN for a length of 10 m on a smooth plane inclined at an angle of 15° with the horizontal is (sin 15° = 0.259)

To find the work done in pulling the block of wood up the inclined plane, we can analyze the forces acting on the block.

Given:
Weight of the block, $W=2\text{kN}$
Length of the inclined plane, $d=10\text{m}$
Angle of inclination with the horizontal, $\theta ={15}^{\circ }$
Sine of the inclination angle, $sin\left({15}^{\circ }\right)\approx 0.2588$ (using the precise value for the options)

Since the inclined plane is smooth, there is no friction opposing the motion. The force required to pull the block up the plane at a constant velocity must balance the component of the weight acting down the incline:

$F=Wsin\left(\theta \right)$

The work done ($U$) is defined as the force multiplied by the displacement along the plane:

$U=F\times d=Wsin\left(\theta \right)\times d$

Substituting the given values into the equation:

$U=2\text{kN}\times 10\text{m}\times sin\left({15}^{\circ }\right)$

Using $sin\left({15}^{\circ }\right)\approx 0.2588$:

$U=2\times 10\times 0.2588\text{kJ}$

$U=5.176\text{kJ}\approx 5.17\text{kJ}$

Therefore, the work done in pulling the block of wood is 5.17 kJ.