A lorry of mass $3.5\times10^4\text{ kg}$ attains a steady speed $\nu$ while climbing an incline of $1$ in $10$ with its engine operating at $175$ kW. Find $\nu$. $g=10ms^{-1}$. Neglect friction.
The answer in the book is $5.0ms^{-1}$.
Attempt 1:
All the power goes into the sum of potential and kinetic energies. $$175000=\frac{1}{2}m\nu^{2}+mgh$$ Relating $h$ to $\nu$ on given slope, $\frac{dh}{dt}=\frac{\nu}{10}$
$$\nu^2 +2\nu -10=0$$ $$\nu=2.32,-4.32 ms^{-1}$$
Not sure what to make of that result.
Attempt 2:
The force component due to gravity perpendicular to the slope is $mg\sin\theta=m=35000N$. The power of the lorry can exert $175000Nms^{-1}$. The resultant force of $140000N$ must go into momentum. $140000=m\nu$. Therefore $\nu=4.0ms^{-1}$.
This attempt gives a nicer value but is not the same as the book. Did I make a calculation mistake? Could the answer in the book be wrong?
The lorry is moving at constant speed so the forces are in equilibrium. Therefore you have $$\frac Pv=mg\sin\theta$$
Plugging in the values, you get $v=5$