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(A) Given: Mass $m = 1000 \; kg$,speed $v = 36 \; km/h = 36 	imes (5/18) = 10 \; m/s$,$	an 	heta = 1/2$,$\mu = 1/\sqrt{3}$.From the triangle,$\sin

Vedclass · Accepted Answer

$94.4 	imes 10^3 \; W$

Vedclass · Answer

(A) Given: Mass $m = 1000 \; kg$,speed $v = 36 \; km/h = 36 	imes (5/18) = 10 \; m/s$,$	an 	heta = 1/2$,$\mu = 1/\sqrt{3}$.From the triangle,$\sin 	heta = 1/\sqrt{5}$ and $\cos 	heta = 2/\sqrt{5}$.The force required to move the engine up the incline at a constant speed is $F = mg \sin 	heta + f_k = mg \sin 	heta + \mu mg \cos 	heta$.$F = mg(\sin 	heta + \mu \cos 	heta) = 1000 	imes 9.8 	imes (1/\sqrt{5} + (1/\sqrt{3}) 	imes (2/\sqrt{5}))$.$F = 9800 	imes (1/\sqrt{5} + 2/\sqrt{15}) = 9800 	imes (0.4472 + 0.5164) \approx 9800 	imes 0.9636 \approx 9443 \; N$.Power $P = F 	imes v = 9443 	imes 10 = 94430 \; W = 94.4 	imes 10^3 \; W$.

An engine of mass $1$ metric ton is ascending an inclined plane,at an angle $\theta = \tan^{-1}(1/2)$ with the horizontal,with a speed of $36 \; km/h$. If the coefficient of friction of the surface is $1/\sqrt{3}$,then the power developed by the engine is:

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