Question

A block of mass $10\text{kg}$ is slowly slid up a smooth inclined plane of inclination ${37}^{\circ }$ by a person. Calculate the work done by the person in moving the block through a distance of $20\text{m}$ along the incline, if the person applies the force parallel to the inclined plane. Take $g=10{\text{m/s}}^2$.

• A. $400\text{J}$
• B. $100\text{J}$
• C. $1600\text{J}$
• D. $1200\text{J}$

Answer 1

The correct option is D $1200\text{J}$

Since person applies force parallel to the incline, and block moves up the inclined plane slowly i.e $v_i=v_f=0$, hence, $\Delta KE=0$ for the block.


Applying work energy theorem on the block for given dispalcement:
$W_{mg}+W_F+W_N=\Delta KE$
$(-mg\times 20\sin {37}^{\circ })+(F\times 20)+0=0$
(-ve sign because $mg$ and vertical displacement of block ($20\sin {37}^{\circ }$) are opposite)


$\Rightarrow 20F=20mg\sin {37}^{\circ }$
$\therefore F=100\times \frac{3}{5}=60\text{N}$
$\Rightarrow$Work done by force $F$:
$W_F=F\times 20=60\times 20=1200\text{J}$