Question

An engine of $1000$kg is moving up an inclined plane $\theta ={30}^{\circ }$ at the rate of $20m{s}^{-1}$. If the coefficient of friction is $\frac{1}{\sqrt{3}}$, power of the engine is:

• A. 98 KW
• B. 196 KW
• C. 392 KW
• D. 49 KW

Answer 1

The correct option is B 196 KW

here ,

$sin\theta =\frac{1}{2}cos\theta =\frac{\sqrt{3}}{2}$

$power=workdone/time=\frac{f.d}{t}=f.vv=velocity$

$f=mgsin\theta +\mu mgcos\theta =mg(sin\theta +\mu cos\theta )$

$f=mg(\frac{1}{2}+\frac{1}{\sqrt{3}}\frac{\sqrt{3}}{2})\mu =\frac{1}{\sqrt{3}}f=mgpower=f.v=mgvbyputtingvalueofm,gandvwegetp=1000\times 9.8\times 20=196000W=196kW$

option B is correct.