Question

The work done in slowly pulling up a block of wood weighing 2kN for a length of 10m on a smooth plane inclined at an angle of ${15}^{\circ }$ with the horizontal by a force parallel to the incline is :? $(sin{15}^{\circ }=0.26,cos{15}^{\circ }=0.96)$

• A. 4.4kJ
• B. 5.2kJ
• C. 8.9kJ
• D. 9.8kJ

Answer 1

The correct option is B 5.2kJ

$\text{Step 1: Work done}$

Let $h$ be the height of block from ground (Refer figure)

Work done = Change in potential energy of block

$\Rightarrow W=mgh$ $....\left(1\right)$

$\text{Step 2: Height of block from ground}$
$h=10\sin {15}^{\circ }$

$\text{Step 3: Substitute the values}$

Weight of block$=mg=2kN$

Put the values of $h$ and $mg$ in equation $\left(1\right)$
$\therefore W=2\times {10}^{3}N\times 1.0\times \sin {15}^{\circ }$$=5.2\times {10}^{3}J=5.2kJ$

Hence Option (B) is correct.