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3 years ago

7

The engine of a locomotive exerts a constant force of 8.1*10^5 N to accelerate a train to 68 km/h. Determine the time (in min) t

aken for the train of mass 1.9*10^7 kg to reach this speed from rest.

2 answers:

Bumek [7]3 years ago

8 0

Answer:443.1 s

Explanation:

Given

Engine of a locomotive exerts a force of $8.1\times 10^5 N$

Mass of train $=1.9\times 10^7$

Final speed (v) $=68 km/h \approx 18.88 m/s$

F=ma

so $acceleration(a) =\frac{F}{m}=\frac{8.1\times 10^5}{1.9\times 10^7}$

$a=0.042631 m/s^2$

and acceleration is

$a=\frac{v-u}{t}$

$0.042631=\frac{18.88-0}{t}$

$t=443.089 \approx 443.1 s$

nalin [4]3 years ago

5 0

Explanation:

It is given that,

Force acting on the engine, $F=8.1\times 10^5\ N$

Initial speed of the engine, u = 0 (at rest)

Final speed of the engine, v = 68 km/h = 18.88 m/s

Mass of the train, $m=1.9\times 10^7\ kg$

We need to find the time taken by the train. Firstly, we will find the acceleration of the engine from Newton's second law of motion as :

$a=\dfrac{F}{m}$

$a=\dfrac{8.1\times 10^5}{1.9\times 10^7}$

$a=0.042\ m/s^2$

Now using first equation of motion to find time taken as :

$v=u+at$

$t=\dfrac{v-u}{a}$

$t=\dfrac{18.88-0}{0.042}$

t = 449.52 seconds

or

t = 7.49 minutes

So, the time taken for the train to reach this speed is 7.49 minutes. Hence, this is the required solution.

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