You're very nearly there.

You're correct to say that $Power = Fv$, and that the velocity at the moment the train reaches the top of the slope is given by $v_\text{top} = 350/(6 + 150g\sin2)$ so the force at the top of the slope is $F_\text{top} = 1000(6 + 150g\sin2)$.

But the acceleration is the *net* force divided by the mass, and the net force is $F_\text{top}$ minus the $6$ kN frictional force i.e.

$$ a = \frac{F_{top} - 6000}{m} = \frac {1000(6 + 150g\sin 2) - 6000}{150000} = g \sin2$$

which using $g = 9.81$ m/sec$^2$ I get as $0.342$ m/sec$^2$.

That's a nice problem - what book is it from? One can get the right answer quickly by projecting forces to the moving axis aligned with the instantaneous direction of V to find d/dt(V), and that is what the book seems to be suggesting (but in a rather obscure way).

Here is a different approach to the solution, probably more straightforward in terms of physics but requiring a bit of algebra and calculus, using coordinates (x,y) on the surface of the wedge; x is the horizontal axis, y is vertical (up the wedge).

There are three forces acting on the body: gravity $mg$, normal reaction $N$, and friction $R$. Since the motion is in the plane of the wedge the sum of forces projected to the normal direction is zero so $N=mg \cos{\theta}$. Within the plane there are two forces. First is the projection of gravity $mg \sin \theta$ directed downward. The second one is the friction force $R$. For $\mu=\tan(\theta)$ a body on inclined plane with the friction coefficient $\mu$ is on threshold of sliding; the standard analysis is given, e.g., in https://en.wikipedia.org/wiki/Friction. Thus the friction force is equal to $R=N \mu = mg \cos{\theta} \tan \theta = mg \sin{\theta}$ and is directed opposite to the velocity vector.

To summarize, the motion within the plane results from action of two forces of equal magnitude $mg \sin{\theta}$, one is directed down the wedge, the other is directed opposite to the velocity vector. From that one can derive that the rate of change of the total velocity magnitude $V$ is equal to that for $V_y$. Initially the body is moving horizontally with velocity $V_0$, so $V=V_0$ and $V_y=0$. Eventually the motion vector is directed down the wedge with $V_f$, so the final values are $V=V_f$ and $V_y=-V_f$. Then it follows that $V_f-V_0=-V_f$, so $V_f=V_0/2$.

The details of the calculation are shown below:

## Best Answer

Hint : You assumed the force to be constant by using $W=Fx$ which is wrong. It is $W=\int Fdx$

Use $P=\frac{dW}{dt}$

Use work energy theorem. Use calculus. Find distance covered when speed is $6ms^{-1}$. That should eliminate your variables if the question has sufficient information.