In Fig. 8-65, a 1400 kgblock of granite is pulled up an incline at a constant speed of 1.34 m/sby a cable and winch. The indicated distances are ${\mathit{d}}_{\mathbf{1}}\mathbf{=}\mathbf{40}\mathbf{}\mathit{m}\mathbf{}\mathbf{and}\mathbf{}{\mathit{d}}_{\mathbf{2}}\mathbf{=}\mathbf{30}$. The coefficient of kinetic friction between the block and the incline is 0.40. What is the power due to the force applied to the block by the cable?

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" ${\mu }_k=0.40$

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

Formula:

$P=Fv\cos \varphi$

From the figure

$$\begin{gathered} \tan \theta =\frac{d_2}{d_1} \\ \Rightarrow \tan \theta =\frac{30}{40} \\ \Rightarrow \tan \theta =0.75 \\ \Rightarrow \theta ={\tan }^{-1}(0.75) \\ \Rightarrow \theta =36.86° \end{gathered}$$ (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.

$$\begin{gathered} F_{net}=T-mg\sin \theta -\mu N \\ \Rightarrow 0=T-mg\sin \theta -\mu mg\cos \theta \\ \Rightarrow T=mg(\sin \theta +\mu \cos \theta ) \\ Power=forceapplied\times velocity \\ \Rightarrow P=T.v \\ \Rightarrow P=Tv\cos \theta \end{gathered}$$ (ii)

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