91Ó°ÊÓ

a) Power capability of the locomotive,$P=1.5MW$

b) Initial speed of the train,$v_i=10m/s$

c) Final speed of the train,$v_f=25m/s$

d) Time interval of acceleration,$t=6.0min$

We use the concept of power and the work-energy principle. From the given work done, we can find the mass of the train. From the equation of power and work done, we can write the equation for speed as a function of time. Also, we can write force as a function of time. By integrating the equation of velocity for a given time, we can find the distance.

Formulae:

The power generated by a work done,$P=\frac{W}{t}$ (i)

The net work done by a body,$W=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$ (ii)

The instantaneous power of a body,$P=Fv$ (iii)

The velocity of a body,$v=\frac{d}{t}$ (iv)

Using equation (i), the work done by the locomotive is given as:

$$\begin{gathered} W=P.t \\ =1.5×{10}^6×360 \\ =5.4×{10}^{8}J \end{gathered}$$

Now, using the given data in equation (ii), we get the mass of the train as:

$$\begin{gathered} 5.4×{10}^8=\frac{1}{2}m{\left(25\right)}^2-\frac{1}{2}m{\left(10\right)}^2 \\ 5.4×{10}^8=\frac{m}{2}\left({\left(25\right)}^2-{\left(10\right)}^2\right) \\ m=\frac{5.4×{10}^8×2}{525} \\ =2.0571×{10}^{6}kg \\ =2.1×{10}^{6}kg \end{gathered}$$

Hence, the mass of the train is $2.1×{10}^{6}kg$.

For any arbitrary t, we can write velocity as a function of t using equations (i) and (ii) as follows:

$$\begin{gathered} Pt=\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_i^2 \\ Pt=\frac{1}{2}m\left(v^2-v_i^2\right) \\ \frac{2Pt}{m}=v^2-v_i^2 \\ v\left(t\right)=\sqrt{v_i^2+\frac{2Pt}{m}} \end{gathered}$$

Plugging the given values, we get the speed of the train as:

$$\begin{gathered} v\left(t\right)=\sqrt{{\left(10\right)}^2\frac{2\left(1.5×{10}^6\right)t}{2.1×{10}^6}} \\ =\sqrt{100+1.5t} \end{gathered}$$

Hence, the speed of the train is $\sqrt{100+1.5t}$.

Force on the train as a function of time using the above speed value and equation (iii) as follows:

$$\begin{gathered} F\left(t\right)=\frac{P}{v\left(t\right)} \\ =\frac{1.5×{10}^6}{\sqrt{100+1.5t}} \end{gathered}$$

Hence, the value of the force is $\frac{1.5×{10}^6}{\sqrt{100+1.5t}}$.

Distance moved by the train during 6.0 min time is given using the value of part (b) in equation (iv) as follows:

$$\begin{gathered} d={∫}_0^{t}v\left(t\right)dt \\ ={∫}_0^{360}\left(\sqrt{100+1.5t}\right)dt \\ ={∫}_0^{360}{\left(100+1.5t\right)}^{\frac{1}{2}}dt \\ ={\left|\frac{{\left(100+1.5t\right)}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}×1.5\right|}_0^{360} \\ ={\left|\frac{{\left(100+1.5t\right)}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle 1.5}}×\left(\frac{1}{1.5}\right)\right|}_0^{360} \\ =6.7×{10}^{3}m \end{gathered}$$

Hence, the value of the distance is $6.7×{10}^{3}m$.