Question

A body of mass $25\text{kg}$ slides down a smooth incline as shown. Find the work done by force of gravity till the body reaches the bottom of the inclined plane ($g=10{\text{ms}}^{-2}$)

• A. $0\text{J}$
• B. $1125\text{J}$
• C. $2250\text{J}$
• D. $4500\text{J}$

Answer 1

The correct option is C $2250\text{J}$

Here for finding the workdone we can take the component of force along the displacement or component of displacement along the direction of force.

Method $\left(\text{I}\right)$
Taking component of force$\left(mg\sin \theta \right)$ along the displacement$\left(s\right)$

Displacement along incline, $|\overrightarrow{s}|=\frac{12}{\cos {37}^{\circ }}$, as $\cos {37}^{\text{o}}=\frac{4}{5}$

$|\overrightarrow{s}|=\frac{12\times 5}{4}=15\text{m}$
Force due to gravity $=mg=25\times 10=250\text{N}$,
Component of force along the incline is $mg\sin \theta =250\times \frac{3}{5}=150\text{N}$
Hence, workdone $W=F\times S=150\times 15=2250\text{J}$, as the force is along the displacement $(\cos 0^{\text{o}}=0)$

Method $\left(\text{II}\right)$
Taking the component of displacement$\left(h\right)$ along force $\left(mg\right)$

By the figure, $\tan {37}^{\text{o}}=\frac{h}{12}$
$\Rightarrow h=12\times \frac{3}{4}=9\text{m}$
Hence work done, $W=F\times h=mg\times h=250\times 9=2250\text{J}$

Option C is the correct answer