An engine of $150$ Kilowatt is drawing a train of total mass $1.5\times 10^5$ Kg up an incline of $1$ in $50$. The frictional resistance is $4$ Kg-wt per ton. Show that the maximum speed of the train is $4\frac{1}{6}$ meters per second.
My attempt:
$150 KW=150\times 100$ W
If $\alpha$ is the inclination of the plane to the horizontal then $\sin{\alpha}=\frac{1}{50}$. Component of weight of the train along the plane=$mg\sin{\alpha}=1.5\times 10^5\times g\times \frac{1}{50}=3000g$ N
Total resistance of the motion $=R=4\times\frac{1.5\times 10^5}{10^3}$ Kg-wt $=600$Kg-wt=$600g$ N (since $1$ ton=$10^3$ Kg)
If F be the force exerted by the train then $F=R+mg\sin{\alpha}=g(600+3000)=3600g$ N
The maximum speed is attained if the acceleration of the train is zero and force exerted by the engine is equal to resistance of the motion.
Let $v$ meter/sec be the required maximum speed of the train.
Please check whether I am on right path and help me to complete the solution of the problem.
Thanks in advance.
Let's first make a free body diagram:
As you said, when the speed is maximum the acceleration is zero, and since the motion is only along the plane and never leaves it, we will be only dealing with the horizontal components: $$\begin{align} P_x+N_x+f_x+F_x&=0\\ -mg\sin(\alpha)-f+F&=0\\ F&=f+mg\sin(\alpha)\tag1 \end{align}$$
$F$ is constant, so let's recall the following formula between the power $P$ of the force $F$ and the work done by it:
$$\begin{align} P&=\frac{\mathrm dW}{\mathrm dt}\\ P&=F\frac{\mathrm dx}{\mathrm dt}\\ P&=Fv_\text{max}\tag2 \end{align}$$
From $(1)$ and $(2)$ we get:
$$\begin{align} P&=(f+mg\sin(\alpha))v_\text{max}\\ v_\text{max}&=\frac P{f+mg\sin(\alpha)}\\ &=\frac{150.10^3}{600\times10+1.5\cdot10^5\times10\times\frac1{50}}\\ &=4.1667=4\frac16 \end{align}$$