P= 2.5 kw, $\mu$ =0.3, $\theta$ =165$^{\circ}$, ν = 2.5 m/s
P=($T_{1}$ – $T_{2}$) ν
2500 = ($T_{1}$ – $T_{2}$) ×2.5 or $T_{1}$ – $T_{2}$ =1000N.
$\frac{T_{1}}{T_{2}}=e^{\mu\theta}=e^{0.3\times 165 \frac{\pi}{180}} $=2.37
or $T_{1}$ = 2.37$T_{2}$ - $T_{2}$ =1000 OR $T_{2}$ =729.9N.
$T_{2}$ =729.9 ×2.37 = 1729.9 N.
Initial Tension, $T_{0}$ = ($T_{1}$ + $T_{2}$)/2 =(1729.9+729.9)/2 = 1229.9N.
(i)When initial tension increase by 8%
$T'_{0}$ = 1229.9 × 1.08 = 1328.3N.
$T_{1}$ + $T_{2}$ = 2656.6
As $\mu$ and $\theta$ remain unchanged, $e^{\mu\theta}$ or $\frac{T_{1}}{T_{2}}$ is same,
2.37 $T_{2}$ + $T_{2}$ =2556.6
$T_{2}$ =788.3N, $T_{1}$ =1868.3N
P = ($T_{1}$ –$T_{2}$) ν = (1868.3-788.3)2.5=2700 W or 2.7 kw.
Increase power = (2.7-2.5)/2.5=0.08 or 8%
(ii) When initial tension is decreased by 8%
$T'_{0}$ =1229.9 × (1-0.08) =1131.5
or $T_{1}$ + $T_{2}$ = 2663
3.37$T_{2}$ = 2263
$T_{2}$ = 671.5N, $T_{1}$ = 1591.5N.
P =(1591.5-671.5)2.5=2300W or 2.3kw.
Decrease in power =(2.5-2.3)/2.5 = 0.08 or 8%
(iii) $\frac{T_{1}}{T_{2}}=e^{\mu\theta}$
$T_{1}$ is the same as before, whereas $\theta$ increases by 8%
1729.9/$T_{2}$ =e$^{0.3}$×((165×1.08$\pi$)/180) =2.54
$T_{2}$ =680.5N.
P =(1729.9-680.5)2.5 =2624W or 2.624kw
Increase in power = (2.624-2.5)/2.5 = 0.0496 or 4.96%
(iv) $\frac{T_{1}}{T_{2}}=e^{\mu\theta}=e^{0.3\times 1.08\times \frac{165\pi}{180}}$=2.54
Or $T_{1}$ = 2.54$T_{2}$
$T_{1}$ + $T_{2}$ = 1229.9 × 2 =2459.8
$T_{2}$ = 694.9 N.
$T_{1}$ =694.9 × 2.54 =1764.9 N.
P = (1764.9-694.9)2.5 = 2675W or 2.675kW
Increase in power = (2.675-2.5)/2.5 =0.07 or 7%