91Ó°ÊÓ

Mass of the block m = 1400 kg

Velocity of the block v = 1.34

Distance$d_1=40\mathrm{m}$

Distance$d_2=30\mathrm{m}$

Coefficient of Friction =role="math" localid="1661403296168" ${\mathrm{μ}}_k=0.40$

By using components of weight, we can find tension in the string. Using value of tension, we can find power.

Formula:

$P=Fv\cos \mathrm{Ï•}$

From the figure

$$\begin{gathered} \tan \mathrm{θ}=\frac{d_2}{d_1} \\ ⇒\tan \mathrm{θ}=\frac{30}{40} \\ ⇒\tan \mathrm{θ}=0.75 \\ ⇒\mathrm{θ}={\tan }^{-1}(0.75) \\ ⇒\mathrm{θ}=36.86° \end{gathered}$$ (i)

As the block is moving with constant velocity $F_{net}=0$

and if T is the tension in the rope then from figure, the force acting along the x axis.

$$\begin{gathered} F_{net}=T-mg\sin \mathrm{θ}-\mathrm{μ}N \\ ⇒0=T-mg\sin \mathrm{θ}-\mathrm{μ}mg\cos \mathrm{θ} \\ ⇒T=mg(\sin \mathrm{θ}+\mathrm{μ}\cos \mathrm{θ}) \\ Power=forceapplied×velocity \\ ⇒P=T.v \\ ⇒P=Tv\cos \mathrm{θ} \end{gathered}$$ (ii)

As tension and velocity are in same direction$\mathrm{θ}=0$using equation (i), (ii) and given values
localid="1663240911359" $$\begin{gathered} P=mg(\sin \mathrm{θ}+\mathrm{μ}\cos \mathrm{θ})v \\ ⇒P=1400×9.8[\sin (36.86)+0.40\cos (36.86\left)\right]×1.34 \\ ⇒P=13720[0.59+0.4×0.8]×1.34 \\ ⇒P=1320[0.59+0.32]×1.34 \\ ⇒P=18384.8×0.91 \\ \mathrm{Solve}\mathrm{further}\mathrm{as}: \\ ⇒P=16729.44 \\ ⇒P≈1.7×{10}^4\mathrm{W} \end{gathered}$$